High School Calculus/Evaluating Limits

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Evaluating Limits[edit]

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[edit]

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[edit]

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.