L'Hospital's Rule: Solving Limits with Infinity - Get Help Now!

[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!

Homework Statement

lim x->infinity <br /> x - ln(1+2e^x)<br />

Homework Equations

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 that's 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!

 

[Dick]

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

 

[needhlpcalc]

...i don't quite follow where you're going with that...can you elaborate more please?

 

[Dick]

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

 

[needhlpcalc]

<br /> x - {ln[(e^x)(2 + 1/e^x)]}<br />

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

cancel out ln(e^x)
<br /> x - x - ln(2 + 1/e^x)<br />

plug in the limit:

<br /> =infinity- 1 - ln(2) <br />

<br /> =infinity<br />

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

 

[heshbon]

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

final solution shoud be ln(2)

 

[needhlpcalc]

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

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

 

[jambaugh]

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.