# Thread: L'Hopital's Rule and Derivatives

1. ## L'Hopital's Rule and Derivatives

I tried solving the following problem many times, but each time I do I end up with the wrong answer. Is the order which I differentiated this expression in correct? Also, is the derivative of e (a constant) e or 0? I'm still very new to calculus, so I'd really appreciate a detailed explanation in regarsds to solving this problem. Thanks.

$\displaystyle 2(1-e^2){\lim_{n \to \infty}(\frac{\frac{e^{2/n}}{n}}{1 - e^{2/n}})$

$= \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{\frac{(n(e^{2/n})' - (e^{2/n})n')}{n^2}}{1' - (\frac{2}{n}(e^{2/n - 1})e')}$

$= \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{\frac{(n(\frac{2}{n}(e^{2/n - 1})e) - (e^{2/n})1)}{n^2}}{-(\frac{2}{n}(e^{2/n - 1})e)}$

$= \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{\frac{(2(e^{2/n}) - (e^{2/n})}{n^2}}{-(\frac{2}{n}(e^{2/n})}$

$= \displaystyle 2(1-e^2){\lim_{n \to \infty}(\frac{(2(e^{2/n}) - (e^{2/n})}{n^2})(\frac{1}{-{2}{n}(e^{2/n})})$

$= \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{2(e^{2/n}) - (e^{2/n})}{n^2(-\frac{2}{n}(e^{2/n}))}$

$= \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{2(e^{2/n}) - (e^{2/n})}{(-2n(e^{2/n}))}$

$= \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{(e^{2/n})(2 - 1)}{-2n(e^{2/n})}$

$= \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{1}{-2n}$

$= \displaystyle 2(1-e^2)(0)$

$= 0$

2. it would be much easier if u wrote it like

$\displaystyle 2(1-e^2) \lim _{n \to \infty} \frac {e^{2/n}}{n(1-e^{2/n})}$

and then apply it

3. Simplify it first. Also, the derivative of any constant is 0. e is no exception.

4. What and when do I simplify?

5. The way yeKciM wrote is the simplified version I was referencing .

6. then just apply the :

$\displaystyle (\frac{u}{v})' = \frac {u'v-v'u}{v^2}$

theres little a bit or work to simplify it after derivation but just go easy

7. Thanks. I'll try that right now.

8. After applying the quotient rule, I ended up with:

$\frac{((n(1 - e^{2/n}))(e^{2/n})' - (e^{2/n})(n(1 - e^{2/n}))'}{(n(1 - e^{2/n}))^2}$

In order to simplify, is it necessary that I find the derivatives of the expressions accompanied by the prime $'$ notation? Or do I ignore the primes and begin by converting this one fraction into two of them?

9. You are making things complicated for yourself.

L'Hopital's rule is used when the limit of your right-hand factor $\displaystyle\huge\frac{\frac{e^{\frac{2}{n}}}{n}} {1-e^{\frac{2}{n}}}$

approaches $\frac{0}{0}$ as n goes to infinity.

It's already in that form, but the differentiation wrt n is inconvenient due to
n being in a denominator position.

Therefore substitute $x=\frac{1}{n}$ and evaluate the limit as x goes to zero instead.

$\displaystyle\lim_{n\rightarrow\infty}\frac{\frac{ e^{\frac{2}{n}}}{n}}{1-e^{\frac{2}{n}}}$

$=\displaystyle\lim_{x\rightarrow\ 0}\frac{xe^{2x}}{1-e^{2x}}$

Now differentiate numerator and denominator since you still have $\frac{0}{0}$
if you try to evaluate the limit.

Use the product rule for the numerator.

10. Correct me if I'm wrong, but I believe there's a problem here. The top, $e^{\frac{2}{n}}$ tends towards 1 as $\to \infty$, whereas the bottom $n(1 - e^{\frac{2}{n}})$ is the indeterminate form $\infty * 0$ as $n \to \infty$?

Let's rewrite the limit as $\frac{\mathop{\lim}\limits_{n \to \infty} e^{\frac{2}{n}}}{ \mathop{\lim}\limits_{n \to \infty} n(1 - e^{\frac{2}{n}})}$

We need to evaluate the bottom first. Change it to

$\mathop{\lim}\limits_{n \to \infty} \frac{1 - e^{\frac{2}{n}}}{\frac{1}{n}}$

Applying L'Hopital's rule,

$\mathop{\lim}{\limits_{n \to \infty} \frac{ \frac{ 2e^{\frac{2}{n}}}{n^2} }{- \frac{1}{n^2}}$.

Which simplifies to $\mathop{\lim}{\limits_{n \to \infty} 2e^{\frac{2}{n}} = -2$.

and so our solution is $-2(1 - e^2) \frac{1}{2} = (e^2 - 1)$?

11. I used the product rule on the numerator and got:

$x(e^{2x})' + (e^{2x})x'$

How do I differentiate the denominator?

12. Math Major, the answer to this problem can be found at Solutions to the Limit Definition of a Definite Integral . The problem was that I didn't understand anything past the L'Hopital's Rule point.

13. Originally Posted by RogueDemon
I used the product rule on the numerator and got:

$x(e^{2x})' + (e^{2x})x'$

How do I differentiate the denominator?
$\frac{d}{dx}\left(1-e^{2x}\right)=0-2e^{2x}$

Now you have $e^{2x}$ as a factor of both numerator and denominator of the result,

so you end up with $-\frac{1}{2}$

finally multiply that by the original expression's LH factor.

14. $\frac{x(e^{2x})' + (e^{2x})x'}{-2(e^{2x})}$

Will this give $\frac{x + x}{-2}$ ?

15. Originally Posted by RogueDemon
$\frac{x(e^{2x})' + (e^{2x})x'}{-2(e^{2x})}$

Will this give $\frac{x + x}{-2}$ ?
No!

u=2x

$\displaystyle\huge\frac{d}{dx}e^{2x}=\frac{d}{dx}e ^u=\frac{du}{dx}\frac{d}{du}e^u=2e^u=2e^{2x}$

then

$\displaystyle\huge\frac{\frac{d}{dx}\left(xe^{2x}\ right)}{\frac{d}{dx}\left(1-e^{2x}\right)}$

$=\displaystyle\huge\frac{x2e^{2x}+e^{2x}(1)}{-2e^{2x}}$

$=\displaystyle\huge\frac{e^{2x}\left(2x+1\right)}{ e^2x(-2)}=\frac{2x+1}{-2}$

Now set x to zero.

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