1. ## [SOLVED] Derivatives

In Doing One Of The Problems, Using L'hopital's Rule,
I Get:

$\displaystyle \frac{xln3 + 3ln3}{xln2}$
But The Simplified form comes to be:
$\displaystyle \frac{ln3}{ln2}$

I don't see what I'm doing wrong in not getting that answer...

2. Originally Posted by >_<SHY_GUY>_<
In Doing One Of The Problems, Using L'hopital's Rule,
I Get:

$\displaystyle \frac{xln3 + 3ln3}{xln2}$
But The Simplified form comes to be:
$\displaystyle \frac{ln3}{ln2}$

I don't see what I'm doing wrong in not getting that answer...

Nothing: both expressions are different. If you write down the original problem perhaps the mistake is before derivating...

Tonio

3. Originally Posted by tonio
Nothing: both expressions are different. If you write down the original problem perhaps the mistake is before derivating...

Tonio
The original problem is is in the attachment, and i stopped where i didn't know what to do next.

Thanks :]

4. Originally Posted by >_<SHY_GUY>_<
The original problem is is in the attachment, and i stopped where i didn't know what to do next.

Thanks :]

Great, but then $\displaystyle \lim_{x\to\infty}\frac{x+3}{x}\cdot \frac{\ln 3}{\ln 2} =\frac{\ln 3}{\ln 2}$ ...what's the problem? It's the limit of a constant ($\displaystyle \frac{\ln 3}{\ln 2}$) times a expression that converges to 1.

Tonio

5. Originally Posted by tonio
Great, but then $\displaystyle \lim_{x\to\infty}\frac{x+3}{x}\cdot \frac{\ln 3}{\ln 2} =\frac{\ln 3}{\ln 2}$ ...what's the problem? It's the limit of a constant ($\displaystyle \frac{\ln 3}{\ln 2}$) times a expression that converges to 1.

Tonio
I Should have seen that...
I struggle alot in calculus, and if it isn't the calculus I struggle with, its the algebra.
Thanks Tonio

6. Originally Posted by >_<SHY_GUY>_<
I Should have seen that...
I struggle alot in calculus, and if it isn't the calculus I struggle with, its the algebra.
Thanks Tonio

Yes...we all did struggle, but as long as it is more beautiful and mind-blowing than hard one continues gladly on the way.

Tonio