Difference between revisions of "Limit of a Function(Definition): Introduction to ε-δ Arguments"

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When most students initially confront the definition above, they are
 
When most students initially confront the definition above, they are
really confused. It helps to ``unravel'' the absolute values a bit.
+
really confused. It helps to "unravel" the absolute values a bit.
 
For example, <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math> is really the same thing as  
 
For example, <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math> is really the same thing as  
  
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On its own, this would mean <math style="vertical-align: 0px">x</math> lies in the <math style="vertical-align: 0px">\delta</math>-neighborhood
 
On its own, this would mean <math style="vertical-align: 0px">x</math> lies in the <math style="vertical-align: 0px">\delta</math>-neighborhood
 
centered at <math style="vertical-align: 0px">c</math>, which is the interval <math style="vertical-align: -5px">(c-\delta,c+\delta)</math>. But
 
centered at <math style="vertical-align: 0px">c</math>, which is the interval <math style="vertical-align: -5px">(c-\delta,c+\delta)</math>. But
what about the other requirement, that <math style="vertical-align: -5px">0<|x-c|</math>? This means we ignore
+
what about the other requirement, that <math style="vertical-align: -5px">0<|x-c|</math>&thinsp;? This means we ignore
 
what happens at <math style="vertical-align: 0px">c</math>.
 
what happens at <math style="vertical-align: 0px">c</math>.
  
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This definition can possibly be better understood through a short
 
This definition can possibly be better understood through a short
video.
+
video (best viewed fullscreen).
 +
<br><br>
  
''< Video coming soon! >''
+
<div style="text-align:center">'''''< Does not work on Internet Explorer >'''''</div>
 +
 
 +
<center><HTML5video width="640" height="420" autoplay="false">LimitDef</HTML5video></center>
 +
 
 +
<br>
  
 
== Proof Approach ==
 
== Proof Approach ==
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== Examples with Linear Functions ==
 
== Examples with Linear Functions ==
  
'''Problem 1.''' Using the definition of a limit, show that <math style="vertical-align: 0px">{\displaystyle \lim_{x\rightarrow1}10=10}</math>.
+
'''Problem 1.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow1}10=10}</math>.
  
''Solution.'' Looking at the statement we need to prove, we have <math style="vertical-align: 0px">f(x)=10,\ L=10</math>
+
''Solution.'' Looking at the statement we need to prove, we have <math style="vertical-align: -5px">f(x)=10,\ L=10</math>
and <math style="vertical-align: 0px">c=1</math>. Since for any <math style="vertical-align: 0px">x</math>, we have <math style="vertical-align: 0px">f(x)=10</math>, we know that for
+
&thinsp;and <math style="vertical-align: -1px">c=1</math>. Since <math style="vertical-align: -5px">f(x)=10</math> for all <math style="vertical-align:0px">x</math>, we know that for any <math style="vertical-align: -4px">x,</math>  
any <math style="vertical-align: 0px">x,</math>  
 
  
 
::<math>|f(x)-L|\,=\,|10-10|\,=\,0\,<\epsilon</math>  
 
::<math>|f(x)-L|\,=\,|10-10|\,=\,0\,<\epsilon</math>  
  
 
as <math style="vertical-align: 0px">\epsilon</math> must be strictly positive. This means any <math style="vertical-align: 0px">\delta</math>
 
as <math style="vertical-align: 0px">\epsilon</math> must be strictly positive. This means any <math style="vertical-align: 0px">\delta</math>
will work. To write it out, you would proceed as follows:
+
will work. To write it out formally, you would proceed as follows:
  
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math> be given. Choose <math style="vertical-align: 0px">\delta=1</math>. Then,
+
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -1px">\delta=1</math>. Then,
whenever <math style="vertical-align: 0px">|x-1|<\delta=1</math>, we have  
+
whenever <math style="vertical-align: -5px">0<|x-c|=|x-1|<\delta=1</math>, we have  
  
 
::<math>|f(x)-L|\ =\ |10-10|\ =\ 0\ <\ \epsilon.</math>
 
::<math>|f(x)-L|\ =\ |10-10|\ =\ 0\ <\ \epsilon.</math>
  
</math>\square</math>
+
<div style="text-align:right"><math>\square</math></div>
 +
A few quick notes about these types of proofs:
 +
 
 +
*Every one will begin with "let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given." That's because most of the time, our <math style="vertical-align: 0px">\delta</math> will be tied to the <math style="vertical-align: 0px">\epsilon</math>.
  
A few quick notes about these.
+
*We conclude the proof with a box/square, indicating we're done.
  
Every one will begin with ``let <math style="vertical-align: 0px">\epsilon>0</math> be given.'' That's
+
<br>
because most of the time, our <math style="vertical-align: 0px">\delta</math> will be tied to the <math style="vertical-align: 0px">\epsilon</math>.
 
  
We conclude the proof with a box/square, indicating we're done.
+
'''Problem 2.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow3}2x-4=2}</math>.
  
'''Problem 2.''' Using the definition of a limit, show that <math style="vertical-align: 0px">{\displaystyle \lim_{x\rightarrow3}2x-4=2}</math>.
+
''Solution.'' In this case, we have <math style="vertical-align: -5px">f(x)=2x-4,\ L=2</math>&thinsp; and <math style="vertical-align: 0px">c=3</math>.
 +
We again begin with scratchwork, and assume our goal. If we <u>''knew''</u>
 +
that <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>, then we would work to get an expression
 +
that has <math style="vertical-align: -5px">|x-c|=|x-3|</math>. It goes like this:
  
''Solution.'' In this case, we have <math style="vertical-align: 0px">f(x)=2x-4,\ L=2</math> and <math style="vertical-align: 0px">c=3</math>.
+
<br>
We again begin with scratchwork, and assume our goal. If we ''knew''
 
that <math style="vertical-align: 0px">|f(x)-L|<\epsilon</math>, then we would work to get an expression
 
that has <math style="vertical-align: 0px">|x-c|=|x-3|</math>. It goes like this:
 
  
 
::<math>\begin{array}{ccccrcl}
 
::<math>\begin{array}{ccccrcl}
Line 102: Line 108:
 
\Rightarrow &  &  &  & |x-3| & < & {\displaystyle \frac{\epsilon}{2}.}
 
\Rightarrow &  &  &  & |x-3| & < & {\displaystyle \frac{\epsilon}{2}.}
 
\end{array}</math>  
 
\end{array}</math>  
 +
<br>
  
This gives us our <math style="vertical-align: 0px">\delta</math>. We will choose a <math style="vertical-align: 0px">\delta<\epsilon/2</math>.  
+
This gives us our <math style="vertical-align: 0px">\delta</math>. We will choose a <math style="vertical-align: -5px">\delta<\epsilon/2</math>.  
  
 
But that was all scratchwork, and the formal writeup looks like a
 
But that was all scratchwork, and the formal writeup looks like a
 
bit different:
 
bit different:
  
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math> be given. Choose <math style="vertical-align: 0px">\delta<\epsilon/2</math>.
+
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -5px">\delta<\epsilon/2</math>.
Then, whenever <math style="vertical-align: 0px">|x-c|=|x-3|<\delta<\epsilon/2</math>, we have  
+
Then, whenever <math style="vertical-align: -5px">0<|x-c|=|x-3|<\delta<\epsilon/2</math>, we have  
 +
 
 +
<br>
  
 
::<math>\begin{array}{ccccrcl}
 
::<math>\begin{array}{ccccrcl}
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\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
 
\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
 
\end{array}</math>
 
\end{array}</math>
 +
<br>
  
 
as required.  
 
as required.  
  
</math>\square</math>
+
<div style="text-align:right;"><math>\square</math></div>
  
It should be mentioned that in cases where <math style="vertical-align: 0px">f(x)=mx+b</math>, and <math style="vertical-align: 0px">m\neq0</math>,
+
It should be mentioned that in cases where <math style="vertical-align: -5px">f(x)=mx+b</math>, and <math style="vertical-align: -5px">m\neq0</math>,
we will get that <math style="vertical-align: 0px">\delta<\epsilon/|m|</math> every time.  
+
we will get that <math style="vertical-align: -5px">\delta<\epsilon/|m|</math> every time.
  
 
== Examples with Quadratic Functions ==
 
== Examples with Quadratic Functions ==
  
'''Problem 3.''' Using the definition of a limit, show that <math style="vertical-align: 0px">{\displaystyle \lim_{x\rightarrow1}x^{2}=1}</math>.
+
'''Problem 3.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow1}x^{2}=1}</math>.
 +
 
 +
''Solution.'' Here, we have <math style="vertical-align: -5px">f(x)=x^{2},\ L=1</math>&thinsp; and <math style="vertical-align: 0px">c=1</math>. We again
 +
begin with scratchwork. Suppose <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>. We then solve
 +
for <math style="vertical-align: -5px">|x-c|=|x-1|</math> to find
  
''Solution.'' Here, we have <math style="vertical-align: 0px">f(x)=x^{2},\ L=1</math> and <math style="vertical-align: 0px">c=1</math>. We again
+
<br>
begin with scratchwork. Suppose <math style="vertical-align: 0px">|f(x)-L|<\epsilon</math>. We then solve
 
for <math style="vertical-align: 0px">|x-c|=|x-1|</math> to find
 
  
 
::<math>\begin{array}{ccccrcl}
 
::<math>\begin{array}{ccccrcl}
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\Rightarrow &  &  &  & |x-1| & < & {\displaystyle \frac{\epsilon}{|x+1|}.}
 
\Rightarrow &  &  &  & |x-1| & < & {\displaystyle \frac{\epsilon}{|x+1|}.}
 
\end{array}</math>  
 
\end{array}</math>  
 +
<br>
  
This certainly describes <math style="vertical-align: 0px">|x-1|</math> in terms of <math style="vertical-align: 0px">\epsilon</math>, but there's
+
This certainly describes <math style="vertical-align: -5px">|x-1|</math> in terms of <math style="vertical-align: 0px">\epsilon</math>, but there's
also an <math style="vertical-align: 0px">x</math> on the right hand side! This requires us to pick an ``initial''
+
also an <math style="vertical-align: 0px">x</math> on the right hand side! This requires us to pick an "initial"
</math>\delta</math>. Let's choose <math style="vertical-align: 0px">\delta=1</math>. Then, whenever <math style="vertical-align: 0px">|x-c|=|x-1|<\delta=1</math>,
+
<math style="vertical-align: 0px">\delta</math>. Let's choose <math style="vertical-align: -1px">\delta=1</math>. Then, whenever <math style="vertical-align: -5px">0<|x-c|=|x-1|<\delta=1</math>,
 
we have  
 
we have  
  
 
::<math>-\delta\,=\,-1\,<\, x-1\,<\,1\,=\,\delta,</math>
 
::<math>-\delta\,=\,-1\,<\, x-1\,<\,1\,=\,\delta,</math>
  
in the manner explained in '''An Explanation'. More importantly, by
+
in the manner explained in '''An Explanation'''. More importantly, by
 
adding two to the inequality we have  
 
adding two to the inequality we have  
  
Line 167: Line 180:
 
::<math>\frac{\epsilon}{3}\,<\,\frac{\epsilon}{x+1}\,=\,\frac{\epsilon}{|x+1|}\,<\,\frac{\epsilon}{1}\,=\,\epsilon.</math>  
 
::<math>\frac{\epsilon}{3}\,<\,\frac{\epsilon}{x+1}\,=\,\frac{\epsilon}{|x+1|}\,<\,\frac{\epsilon}{1}\,=\,\epsilon.</math>  
  
This means that for any <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: 0px">|x-1|<1</math>, we know that <math style="vertical-align: 0px">\epsilon/3<\epsilon/|x+1|</math>.
+
This means that for any <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: -5px">|x-1|<1</math>, we know that <math style="vertical-align: -5px">\epsilon/3<\epsilon/|x+1|</math>.
Thus, we can choose a <math style="vertical-align: 0px">\delta<\epsilon/3</math>, and the proof should work.
+
Thus, we can choose a <math style="vertical-align: -5px">\delta<\epsilon/3</math>, and the proof should work.
There's a small problem, though - we already chose a <math style="vertical-align: 0px">\delta=1</math>.
+
There's a small problem, though - we already chose a <math style="vertical-align: -1px">\delta=1</math>.
The way around this is to use the minimum function.
+
The way around this is to use the '''minimum function''':
  
::When we write <math style="vertical-align: 0px">\min\{a,b\}</math>, it means to take whichever is the
+
<blockquote>When we write <math style="vertical-align: -5px">\min\{a,b\}</math>, it means to take whichever is the
least of both <math style="vertical-align: 0px">a</math> and <math style="vertical-align: 0px">b</math>.  
+
least of both <math style="vertical-align: 0px">a</math> and <math style="vertical-align: 0px">b</math>. </blockquote>
  
 
Now, our proof can be written.
 
Now, our proof can be written.
  
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math> be given. Choose <math style="vertical-align: 0px">\delta=\min\{1,\epsilon/3\}</math>.
+
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -5px">\delta=\min\{1,\epsilon/3\}</math>.
Then, whenever <math style="vertical-align: 0px">|x-c|=|x-1|<\delta</math>,  
+
Then, whenever <math style="vertical-align: -5px">0<|x-c|=|x-1|<\delta</math>,  
 +
 
 +
<br>
  
 
::<math>\begin{array}{ccrcrclccccc}
 
::<math>\begin{array}{ccrcrclccccc}
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\Rightarrow &  &  &  & 3|x-1| & < & \epsilon\\
 
\Rightarrow &  &  &  & 3|x-1| & < & \epsilon\\
 
\\
 
\\
\Rightarrow &  & |x+1||x-1| & < & 3|x-1| & < & \epsilon &  &  &  &  & \textrm{(using }\dagger)\\
+
\Rightarrow &  & |x+1||x-1| & < & 3|x-1| & < & \epsilon &  &  &  &  & \textrm{(using~}\dagger)\\
 
\\
 
\\
 
\Rightarrow &  &  &  & |x^{2}-1| & < & \epsilon\\
 
\Rightarrow &  &  &  & |x^{2}-1| & < & \epsilon\\
Line 192: Line 207:
 
\end{array}</math>
 
\end{array}</math>
  
 +
<br>
 
as required.  
 
as required.  
  
</math>\square</math>
+
<div style="text-align:right;"><math>\square</math></div>
 +
 
 +
<br>
  
'''Problem 4.''' Using the definition of a limit, show that <math style="vertical-align: 0px">{\displaystyle \lim_{x\rightarrow2}3x^{2}=12}</math>.
+
'''Problem 4.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow2}3x^{2}=12}</math>.
  
''Solution.'' This time, <math style="vertical-align: 0px">f(x)=3x^{2},\ L=12</math> and <math style="vertical-align: 0px">c=2</math>. We follow
+
''Solution.'' This time, <math style="vertical-align: -5px">f(x)=3x^{2},\ L=12</math>&thinsp; and <math style="vertical-align: 0px">c=2</math>. We follow
the same pattern, doing the scratchwork first. Assume <math style="vertical-align: 0px">|f(x)-L|<\epsilon</math>.
+
the same pattern, doing the scratchwork first. Assume <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>.
 
Then
 
Then
 +
 +
<br>
  
 
::<math>\begin{array}{ccccrcl}
 
::<math>\begin{array}{ccccrcl}
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\end{array}</math>  
 
\end{array}</math>  
  
Based on the previous problem, let's choose an initial <math style="vertical-align: 0px">\delta=1</math>.
+
<br>
 +
 
 +
Based on the previous problem, let's choose an initial <math style="vertical-align: -1px">\delta=1</math>.
 
Then we have  
 
Then we have  
  
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::<math>\frac{\epsilon}{5}\,<\,\frac{\epsilon}{|x+2|}\,<\,\frac{\epsilon}{3}.</math>
 
::<math>\frac{\epsilon}{5}\,<\,\frac{\epsilon}{|x+2|}\,<\,\frac{\epsilon}{3}.</math>
  
So we can choose <math style="vertical-align: 0px">\delta=\min\left\{ 1,\frac{1}{3}\cdot\frac{\epsilon}{3}\right\} =\min\left\{ 1,\frac{\epsilon}{9}\right\}.</math>  
+
So we can choose <math style="vertical-align: -15px">\delta=\min\left\{ 1,\frac{1}{3}\cdot\frac{\epsilon}{5}\right\} =\min\left\{ 1,\frac{\epsilon}{15}\right\}.</math>  
 
Then the proof works.
 
Then the proof works.
  
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math> be given. Choose <math style="vertical-align: 0px">\delta=\min\left\{ 1,\frac{\epsilon}{9}\right\}</math>.
+
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -16px">\delta=\min\left\{ 1,\frac{\epsilon}{15}\right\}</math>.
Then, whenever <math style="vertical-align: 0px">|x-c|=|x-2|<\delta</math>,  
+
Then, whenever <math style="vertical-align: -5px">0<|x-c|=|x-2|<\delta</math>,  
 +
 
 +
<br>
  
 
::<math>\begin{array}{ccrcrclccccc}
 
::<math>\begin{array}{ccrcrclccccc}
  &  &  &  & |x-2| & < & {\displaystyle \frac{\epsilon}{9}}\\
+
  &  &  &  & |x-2| & < & {\displaystyle \frac{\epsilon}{15}}\\
 +
\\
 +
\Rightarrow &  &  &  & 15|x-2| & < & \epsilon\\
 
\\
 
\\
\Rightarrow &  & & & 9|x-2| & < & \epsilon\\
+
\Rightarrow &  & 3|x+2||x-2| & < & 15|x-2| & < & \epsilon &  &  &  &  & \textrm{(using~}\dagger\dagger)\\
 
\\
 
\\
\Rightarrow &  & 3|x+2||x-2| & < & 9|x-2| & < & \epsilon &  &  &  &  & \textrm{(using }\dagger\dagger)\\
+
\Rightarrow &  &   3|x^{2}-4|& & & < & \epsilon\\
 
\\
 
\\
\Rightarrow &  & &  & |3x^{2}-12| & < & \epsilon\\
+
\Rightarrow &  &   |3x^{2}-12|& & & < & \epsilon\\
 
\\
 
\\
 
\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
 
\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
 
\end{array}</math>
 
\end{array}</math>
 +
<br>
  
 
as required.
 
as required.
  
</math>\square</math>
+
<div style="text-align:right;"><math>\square</math></div>
  
'''Problem 5.'''Using the definition of a limit, show that <math style="vertical-align: 0px">{\displaystyle \lim_{x\rightarrow2}x^{2}+3x+1=11}</math>.
+
<br>
  
''Solution.'' Here, we have <math style="vertical-align: 0px">f(x)=x^{2}+3x+1,\ L=11</math> and <math style="vertical-align: 0px">c=2</math>.
+
'''Problem 5.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow2}x^{2}+3x+1=11}</math>.
We again begin with scratchwork, assuming <math style="vertical-align: 0px">|f(x)-L|<\epsilon</math>. Then
+
 
 +
''Solution.'' Here, we have <math style="vertical-align: -5px">f(x)=x^{2}+3x+1,\ L=11</math>&thinsp; and <math style="vertical-align: 0px">c=2</math>.
 +
We again begin with scratchwork, assuming <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>. Then
 +
 
 +
<br>
  
 
::<math>\begin{array}{ccccrcl}
 
::<math>\begin{array}{ccccrcl}
Line 264: Line 295:
 
\end{array}</math>
 
\end{array}</math>
  
Now, we again assume <math style="vertical-align: 0px">\delta=1</math>, so  
+
Now, we again assume <math style="vertical-align: -1px">\delta=1</math>, so  
  
::<math>-\delta\,=\,-1\,<\, x-2\,<\,1\,=\,\delta,\qquad\qquad\qquad(\natural)</math>
+
::<math>-\delta\,=\,-1\,<\, x-2\,<\,1\,=\,\delta,</math>
  
and adding 7 to the inequality,
+
and adding seven to the inequality,
  
::<math>-\delta\,=\,6\,<\, x+5\,<\,1\,=\,7.</math>
+
::<math>-\delta\,=\,6\,<\, x+5\,<\,1\,=\,7.\qquad\qquad\qquad(\natural)</math>
  
 
Again, dividing <math style="vertical-align: 0px">\epsilon</math> by we have
 
Again, dividing <math style="vertical-align: 0px">\epsilon</math> by we have
Line 276: Line 307:
 
::<math>\frac{\epsilon}{7}\,<\,\frac{\epsilon}{|x+5|}\,<\,\frac{\epsilon}{5}</math>  
 
::<math>\frac{\epsilon}{7}\,<\,\frac{\epsilon}{|x+5|}\,<\,\frac{\epsilon}{5}</math>  
  
for all <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: 0px">|x-2|<1.</math> We can now write the proof.
+
for all <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: -5px">|x-2|<1.</math> We can now write the proof.
  
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math> be given. Choose <math style="vertical-align: 0px">\delta=\min\left\{ 1,\frac{\epsilon}{7}\right\}</math>.  
+
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -16px">\delta=\min\left\{ 1,\frac{\epsilon}{7}\right\}</math>.  
Then, whenever <math style="vertical-align: 0px">|x-c|=|x-2|<\delta,</math>  
+
Then, whenever <math style="vertical-align: -5px">0<|x-c|=|x-2|<\delta,</math>  
  
 
::<math>\begin{array}{ccrcrclccccl}
 
::<math>\begin{array}{ccrcrclccccl}
Line 286: Line 317:
 
\Rightarrow &  &  &  & 7|x-2| & < & \epsilon\\
 
\Rightarrow &  &  &  & 7|x-2| & < & \epsilon\\
 
\\
 
\\
\Rightarrow &  & |x+5||x-2| & < & 7|x-2| & < & \epsilon &  &  &  &  & \textrm{(using }\natural)\\
+
\Rightarrow &  & |x+5||x-2| & < & 7|x-2| & < & \epsilon &  &  &  &  & \textrm{(using~}\natural)\\
 
\\
 
\\
\Rightarrow &  &  &  & |x^{2}+3x-10| & < & \epsilon\\
+
\Rightarrow &  & |x^{2}+3x-10| & & & < & \epsilon\\
 
\\
 
\\
\Rightarrow &  &  &  & |x^{2}+3x+1-11 & < & \epsilon\\
+
\Rightarrow &  &  |x^{2}+3x+1-11|& & & < & \epsilon\\
 
\\
 
\\
 
\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
 
\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
 
\end{array}</math>
 
\end{array}</math>
 +
<br>
  
 
as required.
 
as required.
  
</math>\square</math>
+
<div style="text-align:right;"><math>\square</math></div>
  
 
== Other Examples ==
 
== Other Examples ==
  
'''Problem 6.''' Using the definition of a limit, show that <math style="vertical-align: 0px">{\displaystyle \lim_{x\rightarrow1}\sqrt{x}=1}</math>.  
+
'''Problem 6.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow1}\sqrt{x}=1}</math>.  
  
''Solution.'' We have <math style="vertical-align: 0px">f(x)=\sqrt{x},\ L=1</math> and <math style="vertical-align: 0px">c=1</math>. We start
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''Solution.'' We have <math style="vertical-align: -5px">f(x)=\sqrt{x},\ L=1</math>&thinsp; and <math style="vertical-align: -1px">c=1</math>. We start
our scratchwork as usual, assuming <math style="vertical-align: 0px">|f(x)-L|<\epsilon</math>. Then we have  
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our scratchwork as usual, assuming <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>. Then we have  
 +
<br><br>
  
 
::<math>\begin{array}{cccrcl}
 
::<math>\begin{array}{cccrcl}
Line 313: Line 346:
 
\Rightarrow &  &  & |x-1| & < & |\sqrt{x}+1|\cdot\epsilon.
 
\Rightarrow &  &  & |x-1| & < & |\sqrt{x}+1|\cdot\epsilon.
 
\end{array}</math>
 
\end{array}</math>
 +
<br>
  
Similar to the quadratic examples, we can set an initial <math style="vertical-align: 0px">\delta=1.</math>
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Similar to the quadratic examples, we can set an initial <math style="vertical-align: -1px">\delta=1.</math>
 
Then
 
Then
  
Line 331: Line 365:
 
::<math>\epsilon\,<\,|\sqrt{x}+1|\,<\,\left(\sqrt{2}+1\right)\epsilon\qquad\mbox{while}\qquad\frac{\epsilon}{\sqrt{2}+1}\,<\,\frac{\epsilon}{|\sqrt{x}+1|}\,<\,\frac{\epsilon}{1}.</math>  
 
::<math>\epsilon\,<\,|\sqrt{x}+1|\,<\,\left(\sqrt{2}+1\right)\epsilon\qquad\mbox{while}\qquad\frac{\epsilon}{\sqrt{2}+1}\,<\,\frac{\epsilon}{|\sqrt{x}+1|}\,<\,\frac{\epsilon}{1}.</math>  
  
for all <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: 0px">|x-1|<1=\delta</math>. This gives us our <math style="vertical-align: 0px">\delta</math>,
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for all <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: -5px">|x-1|<1=\delta</math>. This gives us our <math style="vertical-align: 0px">\delta</math>,
 
and we can write the proof.
 
and we can write the proof.
  
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math> be given. Choose <math style="vertical-align: 0px">\delta=\min\left\{ 1,\epsilon\right\}</math>.  
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'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -5px">\delta=\min\left\{ 1,\epsilon\right\}</math>.  
Then, whenever <math style="vertical-align: 0px">|x-c|=|x-1|<\delta</math>,
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Then, whenever <math style="vertical-align: -5px">|x-c|=|x-1|<\delta</math>,
 +
 
 +
<br>
  
 
::<math>\begin{array}{cccrclcc}
 
::<math>\begin{array}{cccrclcc}
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\Rightarrow &  &  & \left|\sqrt{x}+1\right|\cdot\left|\sqrt{x}-1\right| & < & \epsilon\\
 
\Rightarrow &  &  & \left|\sqrt{x}+1\right|\cdot\left|\sqrt{x}-1\right| & < & \epsilon\\
 
\\
 
\\
\Rightarrow &  &  & \left|\sqrt{x}-1\right| & < & \frac{\epsilon}{\left|\sqrt{x}+1\right|} & < & \epsilon,
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\Rightarrow &  &  & \left|\sqrt{x}-1\right| & < & \displaystyle{\frac{\epsilon}{\left|\sqrt{x}+1\right|}} & < & \epsilon,
 
\end{array}</math>
 
\end{array}</math>
 +
<br>
  
as required. Notice that we used both of the inequalities involving
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as required. Notice that we used both of the inequalities involving <math style="vertical-align: 0px">\epsilon</math> and <math style="vertical-align: -10px">\left|\sqrt{x}+1\right|</math>&thinsp; to complete the proof.
</math>\epsilon</math> and <math style="vertical-align: 0px">\left|\sqrt{x}+1\right|</math> to complete the proof.
 
  
</math>\square</math>
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<div style="text-align:right;"><math>\square</math></div>
  
There are many more difficult examples, but these are meant as an
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<div style="text-align:center"> ''There are many more difficult examples, but these are meant as an introduction.''</div>
introduction.  
 
\end{document}
 

Latest revision as of 14:09, 27 June 2017

Formal Definition

We say is the limit of at if, for any , there exists a   such that whenever ,

An Explanation

When most students initially confront the definition above, they are really confused. It helps to "unravel" the absolute values a bit. For example, is really the same thing as

If we then add to each term, we arrive at

In other words, we are trying to restrict to the -neighborhood centered at , which is the interval . Similarly, we can rewrite to become

On its own, this would mean lies in the -neighborhood centered at , which is the interval . But what about the other requirement, that  ? This means we ignore what happens at .

This neighborhood - the interval , minus the point - is known as a punctured neighborhood.

This definition can possibly be better understood through a short video (best viewed fullscreen).

< Does not work on Internet Explorer >
<HTML5video width="640" height="420" autoplay="false">LimitDef</HTML5video>


Proof Approach

The goal in these proofs is always the same: we need to find a , which will usually be expressed in terms of an arbitrary . For example, we might have to choose a , or a , or even a . But in each case, we usually build our proof backwards in what we can refer to as scratchwork.

Scratchwork begins by assuming our desired result, that

From here, we do whatever it takes (usually factoring) to change into . Once we have a statement of the form

this allows us to pick a that will work. We then just reverse the chain of equalities in our scratchwork to construct the proof. It's easier to see in a few examples.

Examples with Linear Functions

Problem 1. Using the definition of a limit, show that .

Solution. Looking at the statement we need to prove, we have  and . Since for all , we know that for any

as must be strictly positive. This means any will work. To write it out formally, you would proceed as follows:

Proof. Let   be given. Choose . Then, whenever , we have

A few quick notes about these types of proofs:

  • Every one will begin with "let   be given." That's because most of the time, our will be tied to the .
  • We conclude the proof with a box/square, indicating we're done.


Problem 2. Using the definition of a limit, show that .

Solution. In this case, we have   and . We again begin with scratchwork, and assume our goal. If we knew that , then we would work to get an expression that has . It goes like this:



This gives us our . We will choose a .

But that was all scratchwork, and the formal writeup looks like a bit different:

Proof. Let   be given. Choose . Then, whenever , we have



as required.

It should be mentioned that in cases where , and , we will get that every time.

Examples with Quadratic Functions

Problem 3. Using the definition of a limit, show that .

Solution. Here, we have   and . We again begin with scratchwork. Suppose . We then solve for to find



This certainly describes in terms of , but there's also an on the right hand side! This requires us to pick an "initial" . Let's choose . Then, whenever , we have

in the manner explained in An Explanation. More importantly, by adding two to the inequality we have

Dividing by this inequality (which reverses its direction), we have

This means that for any satisfying , we know that . Thus, we can choose a , and the proof should work. There's a small problem, though - we already chose a . The way around this is to use the minimum function:

When we write , it means to take whichever is the least of both and .

Now, our proof can be written.

Proof. Let   be given. Choose . Then, whenever ,



as required.


Problem 4. Using the definition of a limit, show that .

Solution. This time,   and . We follow the same pattern, doing the scratchwork first. Assume . Then



Based on the previous problem, let's choose an initial . Then we have

Now adding 4 to the inequality, we have

Dividing by this inequality, we have

So we can choose Then the proof works.

Proof. Let   be given. Choose . Then, whenever ,



as required.


Problem 5. Using the definition of a limit, show that .

Solution. Here, we have   and . We again begin with scratchwork, assuming . Then


Now, we again assume , so

and adding seven to the inequality,

Again, dividing by we have

for all satisfying We can now write the proof.

Proof. Let   be given. Choose . Then, whenever


as required.

Other Examples

Problem 6. Using the definition of a limit, show that .

Solution. We have   and . We start our scratchwork as usual, assuming . Then we have


Similar to the quadratic examples, we can set an initial Then

and adding one to each term we find

This, in turn, means that

Multiplying the inequality by , we have

for all satisfying . This gives us our , and we can write the proof.

Proof. Let   be given. Choose . Then, whenever ,



as required. Notice that we used both of the inequalities involving and   to complete the proof.

There are many more difficult examples, but these are meant as an introduction.