# Thread: L'Hospital's Rule Example with e

1. ## L'Hospital's Rule Example with e

$\lim x\rightarrow 0[\dfrac{e^{9x} - 1 - 9x}{x^{2}}]$

$\lim x\rightarrow 0[\dfrac{e^{9(0)} - 1 - 9(0)}{(0)^{2}}]$

$\lim x\rightarrow 0[\dfrac{e^{0} - 1 - 0}{0}]$

$\lim x\rightarrow 0[\dfrac{1 - 1 - 0}{0} = \dfrac{0}{0}]$ - Indeterminate

$\lim x\rightarrow 0[\dfrac{[(x^{2})(9e^{9x} - 9)] - [(e^{9x} - 1 - 9x)(2x)]}{[(x^{2})^{2}]} ]$ - Using L' Hospital's rule by way of quotient rule

$\lim x\rightarrow 0[\dfrac{[x^{2}9e^{9x}-9x^{2}] - [(2xe^{9x} - 2x - 18x)(2x)]}{x^{4}}$ Algebra right so far?

2. ## Re: L'Hospital's Rule Example with e

Originally Posted by Jason76
$\lim x\rightarrow 0[\dfrac{e^{9x} - 1 - 9x}{x^{2}}]$

$\lim x\rightarrow 0[\dfrac{e^{9(0)} - 1 - 9(0)}{(0)^{2}}]$

$\lim x\rightarrow 0[\dfrac{e^{0} - 1 - 0}{0}]$

$\lim x\rightarrow 0[\dfrac{1 - 1 - 0}{0} = \dfrac{0}{0}]$ - Indeterminate

$\lim x\rightarrow 0[\dfrac{[(x^{2})(9e^{9x} - 9)] - [(e^{9x} - 1 - 9x)(2x)]}{[(x^{2})^{2}]} ]$ - Using L' Hospital's rule by way of quotient rule
You don't differentiate the entire function to use L'Hospital's Rule, you differentiate just the top and bottom. If the limit is indeterminate of the form \displaystyle \begin{align*} \frac{0}{0} \end{align*} or \displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}, then \displaystyle \begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \end{align*}.

3. ## Re: L'Hospital's Rule Example with e

Originally Posted by Prove It
You don't differentiate the entire function to use L'Hospital's Rule, you differentiate just the top and bottom. If the limit is indeterminate of the form \displaystyle \begin{align*} \frac{0}{0} \end{align*} or \displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}, then \displaystyle \begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \end{align*}.
Well, that will save a lot of time.

$\lim x\rightarrow 0[\dfrac{e^{9x} - 1 - 9x}{x^{2}}]$

$\lim x\rightarrow 0[\dfrac{e^{9(0)} - 1 - 9(0)}{(0)^{2}}]$

$\lim x\rightarrow 0[\dfrac{e^{0} - 1 - 0}{0}]$

$\lim x\rightarrow 0[\dfrac{1 - 1 - 0}{0} = \dfrac{0}{0}]$ - Indeterminate

$\lim x\rightarrow 0[\dfrac{e^{9x} - 1 - 9x}{x^{2}}]$ - looking at the original limit again

$\lim x\rightarrow 0[\dfrac{9e^{9x} -9}{2x}]$ - simple as pie Using L' Hospital's Rule

$\lim x\rightarrow 0[\dfrac{9e^{9(0)} -9}{2(0)}]$

$\lim x\rightarrow 0[\dfrac{9 - 9}{0}] = \dfrac{0}{0}$ - Again indeterminate

$\lim x\rightarrow 0[\dfrac{9e^{9x} -9}{2x}]$ - Looking at first derivative again

$\lim x\rightarrow 0[\dfrac{9(9)e^{9x}}{2}]$

$\lim x\rightarrow 0[\dfrac{(18e^{9x}}{2}]$

$\lim x\rightarrow 0[\dfrac{(18e^{9(0)}}{2}] = 9$ Look right as final answer? Computer says no.

4. ## Re: L'Hospital's Rule Example with e

Originally Posted by Jason76
$\lim x\rightarrow 0[\dfrac{9(9)e^{9x}}{2}]$

$\lim x\rightarrow 0[\dfrac{(18e^{9x}}{2}]$
What is 9 times 9 again?

-Dan