# High School Calculus/Evaluating Limits

### Evaluating Limits

What is a limit? A limit is a place on the graph that the function either does not touch or go past.

When evaluating a limit we may have to factor sometimes in order to get L. L is the point in which the function does not touch or go past.

$\lim _{x\to c}f(x)=L$ Let's start off with a rather simple limit.

$\lim _{x\to 3}x^{2}+x+3$ $3^{2}+3+3=15$ $L=15$ As you can see what we did was just plug 3 into the function to get L

This doesn't always work. This is easily shown in fractions.

I will show you two different ways to evaluate the limits. The first is by factoring and the second is by using L'Hopital's rule.

### Evaluating Limits by Factoring

This is a fairly simply concept, not something easily done. It is especially hard if you have a hard time identifying how polynomials can be rewritten.

Ex.1

$\lim _{x\to -2}{\frac {(x+2)^{2}}{x+2}}$ This gives us $L={\frac {0}{0}}$ This is an indeterminate form. This means we have to find some other way to evaluate the limit so we can get the correct L

Let's look at how $(x+2)^{2}$ is factored

By factoring we now get $\lim _{x\to -2}{\frac {(x+2)(x+2)}{x+2}}$ $\lim _{x\to -2}{\frac {1*(x+2)}{1}}$ $-2+2=0$ $L=0$ Ex.2

$\lim _{x\to 2}{\frac {x^{2}+2x-8}{x-2}}$ ${\frac {2^{2}+2*2-8}{2-2}}$ $={\frac {0}{0}}$ Once again this is an indeterminate form. Let's see if we can use factoring to get and answer.

Factoring the polynomial $x^{2}+2x-8$ we find that it equals $(x-2)(x+4)$ Let's use the factored in the limit equation.

$\lim _{x\to 2}{\frac {(x-2)(x+4)}{x-2}}$ As you can see the (x-2) will cancel each other out. Leaving us with

$\lim _{x\to 2}x+4$ $2+4=6$ $L=6$ This type evaluating limits will take some time, but with practice can be done quickly.

### L'Hopital's Rule

This rule is my favorite way to solve limits with indeterminate form.

This way is a bit more advanced so I will cover it briefly, but I will show some examples and the idea behind it. This is probably something you will learn in Calculus II

When you have a limit that you have confirmed that is in indeterminate form you can use L'Hopital's Rule.

This is the rule

When $\lim _{x\to c}f(x)={\frac {0}{0}}$ , ${\frac {\infty }{\infty }}$ , ${\frac {\infty }{0}}$ , ${\frac {-\infty }{\infty }}$ , ${\frac {\infty }{-\infty }}$ , or ${\frac {-\infty }{-\infty }}$ use L'Hopital's rule. Which is $\lim _{x\to c}{\frac {f^{\prime }(x)}{g^{\prime }(x)}}$ Ex. 1

$\lim _{x\to 5}{\frac {x^{2}-3x-10}{x-5}}$ ${\frac {5^{2}-3*5-10}{5-5}}={\frac {0}{0}}$ Now that we have identified that it is in an indeterminate form we use L'Hopital's rule

$\lim _{x\to 5}{\frac {2x-3}{1}}$ $2*5-3=7$ $L=7$ This is an extremely simplified form of how this rule is used. It is a really nice way to solve limit problems that give you indeterminate forms.