# Thread: differentiation of 1n (3x + e^x)

1. ## differentiation of 1n (3x + e^x)

Hi

So I am applying l'Hospital's Rule to

lim (9x)/(ln (3x + e^x))
x-> infinity

What I wondering is ln (3x + e^x) = ln 3x + ln e^x and then do they both just become 1/3x and 1/e^x?

Any insight would be appreciated
Thanks

Still a calculus beginner

2. Originally Posted by calcbeg
Hi

So I am applying l'Hospital's Rule to

lim (9x)/(ln (3x + e^x))
x-> infinity

What I wondering is ln (3x + e^x) = ln 3x + ln e^x and then do they both just become 1/3x and 1/e^x?

Any insight would be appreciated
Thanks

Still a calculus beginner
$\displaystyle ln (3x + e^x) \neq ln 3x + ln e^x$

3. okay so how do I solve ln (3x +e^x)? what is the derivative?

Calculus beginner

4. Originally Posted by calcbeg
okay so how do I solve ln (3x +e^x)? what is the derivative?

Calculus beginner
$\displaystyle \frac{d}{dx}(ln(u))=\frac{du}{u}$

5. Originally Posted by calcbeg
okay so how do I solve ln (3x +e^x)? what is the derivative?

Calculus beginner
Using the chain rule:

Let $\displaystyle 3x + e^x = u$

Then $\displaystyle ln(3x+e^x) = ln(u)$

Now: $\displaystyle \frac{d}{dx}ln(u) = \frac{d}{du}ln(u) \times \frac{du}{dx} = \frac{1}{u} \times (3 + e^x) = \frac{3 + e^x}{3x + e^x}$

6. So lim 9x/ ln (3x + e^x) becomes 9/ (3 + e^x)/(3x + e^x)
x->infinity

which becomes (27x + 9e^x)/(3 + e^x) which is again infinity/infinity

so I am thinking I need to apply the l'Hospital's rule again

so now I apply the quotient rule so it becomes

((27 + 9e^x)(e^x) - (e^x)(27x + 9e^x))/((e^x)^2)

so (27 + 9e^x - 27x-9e^x)/e^x = (27 - 27x)/e^x so I still have infinity/infinity

where am I going wrong????

Thanks

Calculus Beginner

7. Sorry I am now confusing myself

Am I right.... applying l'Hospital's rule a second time I get

lim (27x + 9e^x)/ (3+ e^x) = lim 27 + 9e^x/e^x
x-> infinity

so then I apply l'hospital's rule a third time to get

lim 9e^x/e^x = 9
x-> infinity

Am I right???

Thanks

Calculus Beginner

8. You have to break it up:

$\displaystyle \lim_{x \to \infty} \frac{27x + 9e^x}{3 + e^x} = \lim_{x \to \infty} \frac{\frac{27x}{x}}{\frac{3}{x} + \frac{e^x}{x}} + \frac{\frac{9e^x}{e^x}}{\frac{3}{e^x} + \frac{e^x}{e^x}}$

You'll find that the first part has a limit of 0 and the second part has a limit of 9. So the overall limit is 9.

9. Originally Posted by calcbeg
Sorry I am now confusing myself

Am I right.... applying l'Hospital's rule a second time I get

lim (27x + 9e^x)/ (3+ e^x) = lim 27 + 9e^x/e^x
x-> infinity

so then I apply l'hospital's rule a third time to get

lim 9e^x/e^x = 9
x-> infinity

Am I right???

Thanks

Calculus Beginner
Yes, this is correct, however there really isn't any need to abuse L'Hospital's so much!

After using it for the first time, simply divide each term by $\displaystyle e^x$ and get:

$\displaystyle \lim_{x \to \infty} \frac{27x + 9e^x}{3 + e^x} = \lim_{x \to \infty} \frac{\frac{27x}{e^x} + \frac{9e^x}{e^x}}{\frac{3}{e^x} + \frac{e^x}{e^x}} = \frac{0 + 9}{0 + 1} = 9$