# 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.

${\displaystyle \lim _{x\to c}f(x)=L}$

Let's start off with a rather simple limit.

${\displaystyle \lim _{x\to 3}x^{2}+x+3}$

${\displaystyle 3^{2}+3+3=15}$

${\displaystyle L=15}$

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

${\displaystyle \lim _{x\to -2}{\frac {(x+2)^{2}}{x+2}}}$

This gives us ${\displaystyle L={\frac {0}{0}}}$

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 ${\displaystyle (x+2)^{2}}$  is factored

By factoring we now get ${\displaystyle \lim _{x\to -2}{\frac {(x+2)(x+2)}{x+2}}}$

${\displaystyle \lim _{x\to -2}{\frac {1*(x+2)}{1}}}$

${\displaystyle -2+2=0}$

${\displaystyle L=0}$

Ex.2

${\displaystyle \lim _{x\to 2}{\frac {x^{2}+2x-8}{x-2}}}$

${\displaystyle {\frac {2^{2}+2*2-8}{2-2}}}$

${\displaystyle ={\frac {0}{0}}}$

Once again this is an indeterminate form. Let's see if we can use factoring to get and answer.

Factoring the polynomial ${\displaystyle x^{2}+2x-8}$  we find that it equals ${\displaystyle (x-2)(x+4)}$

Let's use the factored in the limit equation.

${\displaystyle \lim _{x\to 2}{\frac {(x-2)(x+4)}{x-2}}}$

As you can see the (x-2) will cancel each other out. Leaving us with

${\displaystyle \lim _{x\to 2}x+4}$

${\displaystyle 2+4=6}$

${\displaystyle L=6}$

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

Ex. 1

${\displaystyle \lim _{x\to 5}{\frac {x^{2}-3x-10}{x-5}}}$

${\displaystyle {\frac {5^{2}-3*5-10}{5-5}}={\frac {0}{0}}}$

Now that we have identified that it is in an indeterminate form we use L'Hopital's rule

${\displaystyle \lim _{x\to 5}{\frac {2x-3}{1}}}$

${\displaystyle 2*5-3=7}$

${\displaystyle L=7}$

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.