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.

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

Let's start off with a rather simple limit.

$\underset{x\to 3}{\lim }{x}^2+x+3$

$3^2+3+3=15$

$L=15$

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

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

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 $\underset{x\to -2}{\lim }\frac{(x+2)(x+2)}{x+2}$

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

$-2+2=0$

$L=0$

Ex.2

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

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

$=\frac{0}{0}$

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.

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

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

$\underset{x\to 2}{\lim }x+4$

$2+4=6$

$L=6$

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

Ex. 1

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

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

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

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

$2*5-3=7$

$L=7$

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.