Math 22 Limits

The Limit of a Function

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

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

Properties of Limits

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

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

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

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

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

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

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

Techniques for Evaluating Limits

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

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

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

Example: Evaluate $\lim _{x\to 2}{\frac {x^{2}-4}{x^{2}-x-2}}=\lim _{x\to 2}{\frac {(x-2)(x+2)}{(x-2)(x+1)}}=\lim _{x\to 2}{\frac {x+2}{x+1}}$. 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 $\lim _{x\to 0}{\frac {{\sqrt {x+4}}-2}{x}}=\lim _{x\to 0}{\frac {{\sqrt {x+4}}-2}{x}}\cdot {\frac {{\sqrt {x+4}}+2}{{\sqrt {x+4}}+2}}=\lim _{x\to 0}{\frac {(x+4)-4}{x({\sqrt {x+4}}+2)}}=\lim _{x\to 0}{\frac {x}{x({\sqrt {x+4}}+2)}}=\lim _{x\to 0}{\frac {1}{{\sqrt {x+4}}+2}}$. 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 $c$ than it approaches from the right of $c$, 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
 $\lim _{x\to c^{-}}f(x)=L$ and $\lim _{x\to c^{+}}f(x)=K$

One-sided Limit is related to unbounded function.

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

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

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

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

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

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

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