Homework Help: Solving problem l'Hospital Rule

1. Nov 19, 2009

Slimsta

1. The problem statement, all variables and given/known data
solve

2. Relevant equations
f(x)/g(x) = f'(x)/g'(x)

3. The attempt at a solution
1. $$\displaystyle \Large \lim _{x\to \infty }(1+3x)^{1/x}$$ = 0
because 1+3x is infinity..
infinity^(1/infinity)
small#/big# = 0
so infinity^0 = 1

2. $$\displaystyle \Large \lim _{x\to 1}\frac{\ln x}{x^2+x-2}$$ = 1/3
(1/x) / (2x+1).....

3. $$\displaystyle \Large \lim _{x\to \infty }\frac{x^2}{e^x}$$ = infinity
cuz if plugging in infinity, i get infinity2/infinity = infinity

im i right so far?

2. Nov 19, 2009

Staff: Mentor

1 and 3 are wrong, but 2 is right. The following are all indeterminate forms, which means that they aren't numbers, and the limits that are in these forms can come out to be anything.
$$[\infty^0],~\left[\frac{0}{0}\right],~\left[\frac{\infty}{\infty}\right],~[\infty - \infty]$$

Use L'Hopital's Rule on #3, and you should see that the limit isn't infinite. #1 takes a fair amount of time to explain, so see if your textbook has an example that's similar. It involves taking logs before you use L'Hopital's Rule.

3. Nov 19, 2009

Slimsta

for #3.. if i plug in e^infinity it says error..
so it should be undefined right?

for #1
the rule states:
"If the expression
$$\frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}$$
has the form 0/0 or infinity/infinity, then l'Hôpital's rule states that
f(x)/g(x) = f'(x)/g'(x)
provided that that second limit exists."

$$[\infty^0],~[\infty - \infty]$$

i know that for
$$[\infty - \infty]$$
it should be 1/x - 1/sinx or something like that

4. Nov 19, 2009

Staff: Mentor

No, not for #3, and besides e^infinity is not a number. I don't know what you're plugging that into, but #3 has a limit. Use L'Hopital's Rule. If necessary, use it again.
L'Hopital's Rule doesn't apply to these indeterminate forms. For the first one, the usual strategy is to let y = your expression, and then take the log of both sides. That will typically give you a product that you can rewrite as a quotient that is one of the two indeterminate forms [0/0] or [infinity/infinity], and then use L'Hopital's Rule. For the second one, you can sometimes rearrange things to get a quotient on which to use L'Hopital's Rule.
Sort of.
$$\lim_{x \rightarrow 0}\left[ 1/x - 1/sin(x)\right]$$ is an indeterminate form of this type.

5. Nov 19, 2009

Slimsta

oh i get it now.
so #1 would be lny = 1/3 ln(1+3x)
and after solving i get 'infinity'

for #3 it is 2/e^infinity ==> 0
so #3 would be 0

im i correct?

and what if i have a question like this:
$$\displaystyle \Large \lim _{x\to \infty}(\sqrt{x^2+x}-x)$$
i get $$[\infty - \infty]$$
now how do i use $$\lim_{x \rightarrow 0}\left[ 1/x - 1/sin(x)\right]$$?

would it be, (sinx-x)/xsinx and now apply the rule f'(x)/g'(x) ?

6. Nov 19, 2009

Staff: Mentor

#1 is still wrong, but #3 is right, so now you have two out of the three of them. I'll talk about #1 at the end of this post.

For this problem
$$\lim _{x\to \infty}(\sqrt{x^2+x}-x)$$
multiply by 1 in the form of [sqrt(x^2 + x) + x]/[sqrt(x^2 + x) + x]. That will give you a quotient that might be in a form so that you can use L'Hopital's Rule.

For #1, let y = (1 + 3x)1/x
Take the log of both sides : ln y = ln[(1 + 3x)1/x]
Your result of 1/3 ln(1 + 3x) is incorrect. What should you get?
Notice that there is no limit process going on. After a while, we'll bring limits back in.

7. Nov 19, 2009

Slimsta

oh that was my mistake.. for #1 i will get 1/x ln(1 + 3x) and since 1/x is 0 that makes the whole thing 0.. so the final answer is 0.

for the question
$$\lim _{x\to \infty}(\sqrt{x^2+x}-x)$$

i get x/(sqrt{x^2+x}+x)
from there if i use the rule, i get 1/(0.5(x2+x)-0.5*(2x+1)+1)
would that equal 0? cuz 1/infinity?

8. Nov 19, 2009

Staff: Mentor

In fact, the limit in your last problem is 1/2,not 0, so sorry to burst your bubble.
You don't need L'Hopital's Rule for this problem.
$$\lim _{x\to \infty}(\sqrt{x^2+x}-x)~=~ \lim _{x\to \infty}(\sqrt{x^2+x}-x)\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}~=~\lim _{x\to \infty}\frac{x}{\sqrt{x^2+x}+x}$$
Factor x^2 out of the two terms in the radical, and bring it out of the radical as x.
$$\lim _{x\to \infty}\frac{x}{x\sqrt{1+1/x}+x}~=~\lim _{x\to \infty}\frac{x}{x(\sqrt{1+1/x}+1)}~=~ 1/2$$

In the next-to-last step you can cancel the x in the numerator with the x in the denominator, leaving you with 1 in the numerator and and expression that approaches 2 in the denominator.

For #1, you're closer, but no cigar, and also the limit is not zero.
If y = ln(1 + 3x)1/x
ln y = (1/x)*ln(1 + 3x)*3 -------chain rule!!!
= (3*ln(1 + 3x))/x

This is true for all reasonable x (i.e., x > -1/3), so it's also true in the limit as x gets large without bound.

So lim lny = lim (3*ln(1 + 3x))/x with the limit taken as x gets large without bound, in which case both numerator and denominator get large without bound, so this is the indeterminate form [infinty/infinity].

See if you can finish this off. Here's what you need to do:
1. Evaluate the limit on the right, keeping in mind that this is the indeterminate form [infinity/infinity]. Call this value A.
2. If lim lny = A, then ln(lim y) = A as well. As long as the function is continuous (the ln function is), you can switch the order of the function and the limit.
3. Since ln (lim y) = A, what is lim y? Keep in mind that what we wanted all along was lim y = lim (1 + 3x)1/x.

9. Nov 19, 2009

Slimsta

how did you get
"ln y = (1/x)*ln(1 + 3x)*3 -------chain rule!!!"?

i got ln y = (1/x)*ln(1 + 3x) = ln(1 + 3x) / x
==> f'(x)/g'(x) ==>(3/(1+3x))/x = 3x/(1+3x)
==> plug in 'infinity'
3infinity / 1+3infinity = 3/3 = 1

or if it would be infinity/infinity which means i have to apply the rule again,
3/3 = 1 anyways..

in yours
ln y = (1/x)*ln(1 + 3x)*3 where is the last 3 coming from?

Last edited: Nov 19, 2009
10. Nov 20, 2009

Staff: Mentor

That should be:
If y = ln(1 + 3x)1/x
ln y = (1/x)*ln(1 + 3x) = [ln(1 + 3x)]/x

So lim lny = lim (ln(1 + 3x))/x with the limit taken as x gets large without bound, but no you never, never "plug in" infinity. That's what limits are for.

This is how you write it so that you aren't plugging in infinity:
$$\lim_{x \to \infty} ln y~=~\lim_{x \to \infty} \frac{ln(1 + 3x)}{x}~~~~\left[\frac{\infty}{\infty} \right]$$
$$\Rightarrow ln(\lim_{x \to \infty} y)~=~\lim_{x \to \infty} \frac{\frac{3}{1 + 3x}}{1}~=~0\right]$$

In the first line above, the $[\infty/\infty]$ notation is just a reminder to myself that I have an indeterminate form of the type I can use L'Hopital's Rule on. Don't clutter up your work with the f'(x)/g'(x) stuff. If you feel you need to include it, put it off to the side somewhere so that it's not in the main flow of your logic.

In the second line I have switched the order of the limit operation and log function, and have applied L'Hopital's Rule -- that's where the chain rule came in.

The second equation above says that the log of the limit I want is zero, so what is the value of the limit? IOW, ln[lim (1 + 3x)1/x)] = 0, so what's the value of this limit?

11. Nov 20, 2009

Slimsta

ln[lim (1 + 3x)1/x)] = 0
==> lim (1 + 3x)1/x)= e^0
==> lim (1 + 3x)1/x)= 1

thanks i got it!
i really appreciate ur help!

12. Nov 20, 2009

Staff: Mentor

Sure, you're welcome. And I appreciate your tenacity. You made a lot of mistakes along the way in this thread, but you hung in there and kept working at it and trying to do better, which is to be admired.