Limit of a Function(Definition): Introduction to ε-δ Arguments
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
- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -\epsilon \,<\,f(x)-L\,<\,\epsilon .}
If we then add to each term, we arrive at
- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L-\epsilon \,<\,f(x)\,<\,L+\epsilon .}
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).
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
- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |f(x)-L|\,<\,\epsilon .}
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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \delta =1} . Then, whenever Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 0<|x-c|=|x-1|<\delta =1} , 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 0<|x-c|=|x-3|<\delta <\epsilon /2} , we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{ccccrcl} & & & & |x-3| & < & {\displaystyle \frac{\epsilon}{2}}\\ \\ \Rightarrow & & & & |2x-6| & < & \epsilon\\ \\ \Rightarrow & & & & |2(x-3)| & < & \epsilon\\ \\ \Rightarrow & & & & |2x-6| & < & \epsilon\\ \\ \Rightarrow & & & & |(2x-4)-2| & < & {\displaystyle \epsilon}\\ \\ \Rightarrow & & & & |f(x)-L| & < & \epsilon, \end{array}}
as required.
It should be mentioned that in cases where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=mx+b} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\neq0} , we will get that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta<\epsilon/|m|} every time.
Examples with Quadratic Functions
Problem 3. Using the definition of a limit, show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \lim_{x\rightarrow1}x^{2}=1}} .
Solution. Here, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=x^{2},\ L=1} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c=1} . We again begin with scratchwork. Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |f(x)-L|<\epsilon} . We then solve for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|=|x-1|} to find
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{ccccrcl} & & & & \left|x^{2}-1\right| & < & \epsilon\\ \\ \Rightarrow & & & & |(x-1)(x+1)| & < & \epsilon\\ \\ \Rightarrow & & & & |x+1|\cdot|x-1| & < & \epsilon\\ \\ \Rightarrow & & & & |x-1| & < & {\displaystyle \frac{\epsilon}{|x+1|}.} \end{array}}
This certainly describes Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-1|} in terms of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} , but there's also an Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} on the right hand side! This requires us to pick an "initial" Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta} . Let's choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=1} . Then, whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|=|x-1|<\delta=1} , we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\delta\,=\,-1\,<\, x-1\,<\,1\,=\,\delta,}
in the manner explained in An Explanation. More importantly, by adding two to the inequality we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1\,<\, x+1\,<\,3.\qquad\qquad\qquad(\dagger)}
Dividing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} by this inequality (which reverses its direction), we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\epsilon}{3}\,<\,\frac{\epsilon}{x+1}\,=\,\frac{\epsilon}{|x+1|}\,<\,\frac{\epsilon}{1}\,=\,\epsilon.}
This means that for any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} satisfying Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-1|<1} , we know that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon/3<\epsilon/|x+1|} . Thus, we can choose a Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta<\epsilon/3} , and the proof should work. There's a small problem, though - we already chose a Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=1} . The way around this is to use the minimum function:
When we write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \min\{a,b\}} , it means to take whichever is the least of both Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} .
Now, our proof can be written.
Proof. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon>0} be given. Choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=\min\{1,\epsilon/3\}} . Then, whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|=|x-1|<\delta} ,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{ccrcrclccccc} & & & & |x-1| & < & {\displaystyle \frac{\epsilon}{3}}\\ \\ \Rightarrow & & & & 3|x-1| & < & \epsilon\\ \\ \Rightarrow & & |x+1||x-1| & < & 3|x-1| & < & \epsilon & & & & & \textrm{(using~}\dagger)\\ \\ \Rightarrow & & & & |x^{2}-1| & < & \epsilon\\ \\ \Rightarrow & & & & |f(x)-L| & < & \epsilon, \end{array}}
as required.
Problem 4. Using the definition of a limit, show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \lim_{x\rightarrow2}3x^{2}=12}} .
Solution. This time, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=3x^{2},\ L=12} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c=2} . We follow the same pattern, doing the scratchwork first. Assume Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |f(x)-L|<\epsilon} . Then
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{ccccrcl} & & & & \left|3x^{2}-12\right| & < & \epsilon\\ \\ \Rightarrow & & & & 3|x^{2}-4| & < & \epsilon\\ \\ \Rightarrow & & & & 3|(x+2|\cdot|x-2)| & < & \epsilon\\ \\ \Rightarrow & & & & |x-2| & < & {\displaystyle \frac{\epsilon}{3|x+2|}.} \end{array}}
Based on the previous problem, let's choose an initial Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=1} . Then we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\delta\,=\,-1\,<\, x-2\,<\,1\,=\,\delta.}
Now adding 4 to the inequality, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3\,<\, x+2\,<\,5.\qquad\qquad\qquad(\dagger\dagger)}
Dividing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} by this inequality, we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\epsilon}{5}\,<\,\frac{\epsilon}{|x+2|}\,<\,\frac{\epsilon}{3}.}
So we can choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=\min\left\{ 1,\frac{1}{3}\cdot\frac{\epsilon}{5}\right\} =\min\left\{ 1,\frac{\epsilon}{15}\right\}.} Then the proof works.
Proof. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon>0} be given. Choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=\min\left\{ 1,\frac{\epsilon}{15}\right\}} . Then, whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|=|x-2|<\delta} ,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{ccrcrclccccc} & & & & |x-2| & < & {\displaystyle \frac{\epsilon}{15}}\\ \\ \Rightarrow & & & & 15|x-2| & < & \epsilon\\ \\ \Rightarrow & & 3|x+2||x-2| & < & 15|x-2| & < & \epsilon & & & & & \textrm{(using~}\dagger\dagger)\\ \\ \Rightarrow & & 3|x^{2}-4|& & & < & \epsilon\\ \\ \Rightarrow & & |3x^{2}-12|& & & < & \epsilon\\ \\ \Rightarrow & & & & |f(x)-L| & < & \epsilon, \end{array}}
as required.
Problem 5. Using the definition of a limit, show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \lim_{x\rightarrow2}x^{2}+3x+1=11}} .
Solution. Here, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=x^{2}+3x+1,\ L=11} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c=2} . We again begin with scratchwork, assuming Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |f(x)-L|<\epsilon} . Then
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{ccccrcl} & & & & \left|x^{2}+3x+1-11\right| & < & \epsilon\\ \\ \Rightarrow & & & & |x^{2}+3x-10| & < & \epsilon\\ \\ \Rightarrow & & & & |(x+5)(x-2)| & < & \epsilon\\ \\ \Rightarrow & & & & |(x+5|\cdot|x-2)| & < & \epsilon\\ \\ \Rightarrow & & & & |x-2| & < & {\displaystyle \frac{\epsilon}{|x+5|}.} \end{array}}
Now, we again assume Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=1} , so
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\delta\,=\,-1\,<\, x-2\,<\,1\,=\,\delta,}
and adding seven to the inequality,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\delta\,=\,6\,<\, x+5\,<\,1\,=\,7.\qquad\qquad\qquad(\natural)}
Again, dividing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} by we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\epsilon}{7}\,<\,\frac{\epsilon}{|x+5|}\,<\,\frac{\epsilon}{5}}
for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} satisfying Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-2|<1.} We can now write the proof.
Proof. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon>0} be given. Choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=\min\left\{ 1,\frac{\epsilon}{7}\right\}} . Then, whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0<|x-c|=|x-2|<\delta,}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{ccrcrclccccl} & & & & |x-2| & < & {\displaystyle \frac{\epsilon}{7}}\\ \\ \Rightarrow & & & & 7|x-2| & < & \epsilon\\ \\ \Rightarrow & & |x+5||x-2| & < & 7|x-2| & < & \epsilon & & & & & \textrm{(using~}\natural)\\ \\ \Rightarrow & & |x^{2}+3x-10| & & & < & \epsilon\\ \\ \Rightarrow & & |x^{2}+3x+1-11|& & & < & \epsilon\\ \\ \Rightarrow & & & & |f(x)-L| & < & \epsilon, \end{array}}
as required.
Other Examples
Problem 6. Using the definition of a limit, show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \lim_{x\rightarrow1}\sqrt{x}=1}} .
Solution. We have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\sqrt{x},\ L=1}
and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c=1}
. We start
our scratchwork as usual, assuming Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |f(x)-L|<\epsilon}
. Then we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{cccrcl} & & & |\sqrt{x}-1| & < & {\displaystyle \epsilon}\\ \\ \Rightarrow & & & |\sqrt{x}+1|\cdot|\sqrt{x}-1 & < & |\sqrt{x}+1|\cdot\epsilon\\ \\ \Rightarrow & & & |x-1| & < & |\sqrt{x}+1|\cdot\epsilon. \end{array}}
Similar to the quadratic examples, we can set an initial Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=1.} Then
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\delta\,=\,-1\,<\, x-1\,<\,1\,=\,\delta,}
and adding one to each term we find
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\,<\, x\,<\,2.}
This, in turn, means that
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1\,<\,\sqrt{x}+1\,<\,\sqrt{2}+1.}
Multiplying the inequality by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} , we have
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}.}
for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} satisfying Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-1|<1=\delta} . This gives us our Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta} , and we can write the proof.
Proof. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon>0} be given. Choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=\min\left\{ 1,\epsilon\right\}} . Then, whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|=|x-1|<\delta} ,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{cccrclcc} & & & |x-1| & < & {\displaystyle \epsilon}\\ \\ \Rightarrow & & & |(\sqrt{x}+1)(\sqrt{x}-1)| & < & \epsilon\\ \\ \Rightarrow & & & \left|\sqrt{x}+1\right|\cdot\left|\sqrt{x}-1\right| & < & \epsilon\\ \\ \Rightarrow & & & \left|\sqrt{x}-1\right| & < & \displaystyle{\frac{\epsilon}{\left|\sqrt{x}+1\right|}} & < & \epsilon, \end{array}}
as required. Notice that we used both of the inequalities involving and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|\sqrt{x}+1\right|} to complete the proof.