## Tuesday, July 14, 2015

### Advanced Math Solutions – Limits Calculator, L’Hopital’s Rule

In the previous posts, we have talked about different ways to find the limit of a function. We have gone over factoring, functions with square roots, and rational functions. What happens when none of those options work? That is when L’Hopital’s Rule comes in.

L’Hopital’s Rule:
If \lim_{x\to a}(\frac{f(x)}{g(x)})=\frac{0}{0}  or  \lim_{x\to a}(\frac{f(x)}{g(x)})=\frac{\pm\infty}{\pm\infty}, where a is finite or \pm\infty,
Then \lim_{x\to a}(\frac{f(x)}{g(x)})=\lim_{x\to a}(\frac{f^{'}(x)}{g^{'}(x)})

What does L’Hopital’s Rule mean?

If f(x) and g(x) are differentiable and if the limit of \frac{f(x)}{g(x)} as x approaches a is \frac{0}{0} or \frac{\pm\infty}{\pm\infty}, then we take the derivative of the numerator and the derivative of the denominator,  \frac{f^{'}(x)}{g^{'}(x)} . The limit of \frac{f(x)}{g(x)} as x approaches a is equal to the limit of \frac{f^{'}(x)}{g^{'}(x)} as x approaches a.

When can you use L’Hopital’s Rule?
1. When the limit is in the indeterminate form of \frac{0}{0} or \frac{\pm\infty}{\pm\infty}
2. Sometimes when the limit is in the other indeterminate forms
3. When the function has an exponent and the limit is an indeterminate form

Here we will work out the first problem step by step (click here):

Try Substitution
We get an indeterminate form of \frac{0}{0}

Take the derivative of the numerator:

Take the derivative of the denominator:

Simplify the function:
\lim _{x\to \:-3}\left(\frac{\frac{x}{\sqrt{x^2+16}}}{1}\right)=\lim _{x\to \:-3}\left(\frac{x}{\sqrt{x^2+16}}\right)

Substitution:
\frac{-3}{\sqrt{\left(-3\right)^2+16}}=-\frac{3}{5}