Math 22 Limits

The Limit of a Function

 Definition of the Limit of a Function
If ${\displaystyle f(x)}$ becomes arbitrarily close to a single number ${\displaystyle L}$ as ${\displaystyle x}$ approaches ${\displaystyle c}$ from either side, then
${\displaystyle \lim _{x\to c}f(x)=L}$
which is read as "the limit of ${\displaystyle f(x)}$ as ${\displaystyle x}$ approaches ${\displaystyle c}$ is ${\displaystyle L}$


Note: Many times the limit of ${\displaystyle f(x)}$ as ${\displaystyle x}$ approaches ${\displaystyle c}$ is simply ${\displaystyle f(c)}$, so limit can be evaluate by direct substitution as ${\displaystyle \lim _{x\to c}f(x)=f(c)}$

Properties of Limits

Let ${\displaystyle b}$ and ${\displaystyle c}$ be real numbers, let ${\displaystyle n}$ be a positive integer, and let ${\displaystyle f}$ and ${\displaystyle g}$ be functions with the following limits ${\displaystyle \lim _{x\to c}f(x)=L}$ and ${\displaystyle \lim _{x\to c}g(x)=K}$. Then

1. Scalar multiple: ${\displaystyle \lim _{x\to c}[bf(x)]=bL}$

2. Sum or difference: ${\displaystyle \lim _{x\to c}[f(x)\pm g(x)]=L\pm K}$

3. Product: ${\displaystyle \lim _{x\to c}[f(x)\cdot g(x)]=L\cdot K}$

4. Quotient: ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}={\frac {L}{K}}}$

5. Power: ${\displaystyle \lim _{x\to c}[f(x)]^{n}=L^{n}}$

6. Radical: ${\displaystyle \lim _{x\to c}{\sqrt[{n}]{f(x)}}={\sqrt[{n}]{L}}}$

Techniques for Evaluating Limits

1. Direct Substitution: Direct Substitution can be used to find the limit of a Polynomial Function.

Example: Evaluate ${\displaystyle \lim _{x\to 3}x^{2}+2x-1=(3)^{2}+2(3)-1=14}$

2. Dividing Out Technique: When direct substitution fails and numerator or/and denominator can be factored.

Example: Evaluate ${\displaystyle \lim _{x\to 2}{\frac {x^{2}-4}{x^{2}-x-2}}=\lim _{x\to 2}{\frac {(x-2)(x+2)}{(x-2)(x+1)}}=\lim _{x\to 2}{\frac {x+2}{x+1}}}$. Now we can use direct substitution to get the answer.

3. Rationalizing (Using Conjugate): When direct substitution fails and either numerator or denominator has a square root. In this case, we can try to multiply both numerator and denominator by the conjugate.

Example: Evaluate ${\displaystyle \lim _{x\to 0}{\frac {{\sqrt {x+4}}-2}{x}}=\lim _{x\to 0}{\frac {{\sqrt {x+4}}-2}{x}}\cdot {\frac {{\sqrt {x+4}}+2}{{\sqrt {x+4}}+2}}=\lim _{x\to 0}{\frac {(x+4)-4}{x({\sqrt {x+4}}+2)}}=\lim _{x\to 0}{\frac {x}{x({\sqrt {x+4}}+2)}}=\lim _{x\to 0}{\frac {1}{{\sqrt {x+4}}+2}}}$. Now we can use direct substitution to get the answer

One-Sided Limits and Unbounded Function

 when a function approaches a different value from the left of ${\displaystyle c}$ than it approaches from the right of ${\displaystyle c}$, the limit does not exists. However, this type of behavior can be described more concisely with
the concept of a one-sided limit. We denote
${\displaystyle \lim _{x\to c^{-}}f(x)=L}$ and ${\displaystyle \lim _{x\to c^{+}}f(x)=K}$


One-sided Limit is related to unbounded function.

In some case, the limit of ${\displaystyle f(x)}$ can be increase/decrease without bound as ${\displaystyle x}$ approaches ${\displaystyle c}$. We can write ${\displaystyle \lim _{x\to c}f(x)=\pm \infty }$

Now, consider ${\displaystyle \lim _{x\to 1}{\frac {-2}{x-1}}}$. By direct substitution, it is of the form ${\displaystyle {\frac {\text{constant}}{0}}}$, so the answer will be either ${\displaystyle \infty }$ or ${\displaystyle -\infty }$. In order to find the limit, we must consider the limit from both side (${\displaystyle \lim _{x\to 1^{-}}}$ and ${\displaystyle \lim _{x\to 1^{+}}}$).

When ${\displaystyle x\to 1^{-}}$, so ${\displaystyle x<1}$, hence ${\displaystyle x-1<0}$. Therefore, ${\displaystyle \lim _{x\to 1^{-}}{\frac {-2}{x-1}}={\frac {\text{negative}}{\text{negative}}}=\infty }$

When ${\displaystyle x\to 1^{+}}$, so ${\displaystyle x>1}$, hence ${\displaystyle x-1>0}$. Therefore, ${\displaystyle \lim _{x\to 1^{+}}{\frac {-2}{x-1}}={\frac {\text{negative}}{\text{positive}}}=-\infty }$

Notice: ${\displaystyle \lim _{x\to 1^{-}}{\frac {-2}{x-1}}\neq \lim _{x\to 1^{+}}{\frac {-2}{x-1}}}$.

So, ${\displaystyle \lim _{x\to 1}{\frac {-2}{x-1}}}$ does not exists