L’Hopital’s Rule for Limits

1. Now we are concerned about limits and we can have nightmarish limits. For example if
(1a)\displaystyle\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0
(1b)\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{0}{0}?
What can we do?

We end up using derivatives to help evaluate such limits. We have L’Hopital’s rule state, if g'(c)\not=0, then
(1c)\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}

2. Sketch of Proof. We can use linear approximations writing
(2)\displaystyle f(x+h)=f(x)+f'(x)h and \displaystyle g(x+h)=g(x)+g'(x)h
Use these approximations in (1a), then take the limit h\to 0. So our limit becomes
(3)\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{h\to0}\lim_{x\to c}\frac{f(x)+f'(x)h}{g(x)+g'(x)h}
We evaluate the inner limit first
(4)\displaystyle\lim_{h\to0}\lim_{x\to c}\frac{f(x)+f'(x)h}{g(x)+g'(x)h}=\lim_{h\to0}\frac{f(c)+f'(c)h}{g(c)+g'(c)h}
But look f(c)=g(c)=0, so (4) simplifies to
(5)\displaystyle  \lim_{h\to0}\frac{f(c)+f'(c)h}{g(c)+g'(c)h}=  \lim_{h\to0}\frac{0+f'(c)h}{0+g'(c)h}
Thus we obtain
(6)\displaystyle  \lim_{h\to0}\frac{f'(c)h}{g'(c)h}=  \lim_{h\to0}\frac{f'(c)}{g'(c)}=\frac{f'(c)}{g'(c)}
So we obtain
(7)\displaystyle  \lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}

3. Sketch of Proof for \infty/\infty. When
(8a)\displaystyle\lim_{x\to c}f(x)=\pm\infty
(8b)\displaystyle\lim_{x\to c}g(x)=\pm\infty
we run into similar problems. L’Hopital’s rule still holds. We just do the following trick:
(9)\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{1/g(x)}{1/f(x)}
and we recover the previous case…kind of!

Taking the derivative of top and bottom, we get
(10)\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{-g'(x)/[g(x)^{2}]}{-f'(x)/[f(x)^{2}]}
We can cross multiply, giving us
(11)\displaystyle  \frac{-g'(x)/[g(x)^{2}]}{-f'(x)/[f(x)^{2}]}=\frac{g'(x)[f(x)^{2}]}{f'(x)[g(x)^{2}]}
Cross multiplying gives us
(12)\displaystyle\lim_{x\to c}\frac{g(x)}{f(x)}=\lim_{x\to c}\frac{g'(x)}{f'(x)}
and thus
(13)\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}.

Notation: We will write
(14)\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}\stackrel{L'H}{=}\lim_{x\to c}\frac{f'(x)}{g'(x)}
when we apply L’Hopital’s rule.

Example 1. Consider the sequence
(15)\displaystyle a_{n}=\frac{n^{2}}{\exp(n)}.
What happens as n\to\infty? We have to apply L’Hopital’s rule twice
(16)\displaystyle \begin{aligned}a_{n}&=\frac{n^{2}}{\exp(n)}\\  &\stackrel{L'H}{=}\frac{2n}{\exp(n)}\\  &\stackrel{L'H}{=}\frac{2}{\exp(n)}\end{aligned}.
Thus we have
(17)\displaystyle \lim_{n\to\infty}\frac{n^{2}}{\exp(n)}\stackrel{L'H}{=}\lim_{n\to\infty}\frac{2}{\exp(n)}=0.


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Calculus, Derivative, Limits, Sequences. Bookmark the permalink.

4 Responses to L’Hopital’s Rule for Limits

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