# L'Hospitals problem NEED HELP

1. Dec 4, 2008

### needhlpcalc

L'Hospitals problem...NEED HELP!!!

This problem is on a test that our teacher said we could research if we wanted...and we are finishing it tomorrow. I have NO CLUE how to approach it and need help!

1. The problem statement, all variables and given/known data
lim x->infinity $$x - ln(1+2e^x)$$

2. Relevant equations

3. The attempt at a solution

so, its infinity - infinity, which is indeterminate.

I did the limit function on my TI-89, and it just keeps spitting the original equation back at me as the answer. I then graphed the function on my TI-89, and when it goes past x=2302, the graph is undefined. HOWEVER, when i go into the table, the values after 2302 go to -infinity, so I think thats what the answer's supposed to be...but I have no idea how to set up the quotient and solve using L'Hospital's Rule...PLEASE HELP!!!

2. Dec 4, 2008

### Dick

Re: L'Hospitals problem...NEED HELP!!!

Don't use l'Hopital on that! Write (1+2e^x)=(e^x)*(2+1/e^x) and use rules of logs.

3. Dec 4, 2008

### needhlpcalc

Re: L'Hospitals problem...NEED HELP!!!

...i dont quite follow where you're going with that...can you elaborate more please?

4. Dec 4, 2008

### Dick

Re: L'Hospitals problem...NEED HELP!!!

It's your turn to elaborate. Simplify ln((e^x)*(2+1/e^x)).

5. Dec 4, 2008

### needhlpcalc

Re: L'Hospitals problem...NEED HELP!!!

$$x - {ln[(e^x)(2 + 1/e^x)]}$$

rule of logs:
$$x - [ln(e^x) + ln(2 + 1/e^x)]$$

cancel out ln(e^x)
$$x - x - ln(2 + 1/e^x)$$

plug in the limit:

$$=infinity- 1 - ln(2)$$

$$=infinity$$

+infinity is the answer? when my calc says -infinity?

Last edited: Dec 4, 2008
6. Dec 4, 2008

### heshbon

Re: L'Hospitals problem...NEED HELP!!!

ln(e^x)=x; not 1.

final solution shoud be ln(2)

7. Dec 4, 2008

### needhlpcalc

Re: L'Hospitals problem...NEED HELP!!!

Oops...dumb mistake. My bad. I feel like a retard. :(

ohhh...ok I think I got it. Thanks to all! :)

Last edited: Dec 4, 2008
8. Dec 4, 2008

### jambaugh

Re: L'Hospitals problem...NEED HELP!!!

To apply l'Hospital's rule recall that since $$e^x$$ is a continuous function you can "move" limits in and out of an exponential.

Thus the limit you want being $$L=\lim_{x\to \infty} f(x) -g(x)$$,

$$\exp({\lim_{x\to \infty} f(x) -g(x))= \lim_{x\to \infty} e^{f(x)-g(x)}$$

Then apply rules of exponentials:
$$\exp(\lim_{x\to \infty} f(x) -g(x))= \lim_{x\to \infty} \frac{e^{f(x)}}{e^{g(x)}}$$

You can now apply l'Hospital's rule on this limit of a quotient and the answer is the exponential of your desired limit. (if it is finite and positive).

Now there may be a more direct way to calculate but this is how you deal with differences of infinities in general so you will want to practice this method.

There is a bit more to it... I should rather have written, given the exponential function is continuous:
$$\lim_{x\to a}e^{h(x)} = \lim_{y\to L} e^y$$ where $$L = \lim_{x\to a} h(x)$$.
But this is the same thing provided the limit L is finite. It just generalizes to the case where L is infinite.

But the result is that for your limit:
$$\lim_{x\to \infty} f(x) -g(x) = \ln\left[ \lim_{x\to \infty} \frac{e^{f(x)}}{e^{g(x)}}\right]$$
provided this logarithm is well defined.