Solving problem l'Hospital Rule

[Slimsta]

Homework Statement

solve

Homework Equations

f(x)/g(x) = f'(x)/g'(x)

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?

 

[Mark44]

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.

 

[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
<br /> \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}<br />
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."

what about
<br /> [\infty^0],~[\infty - \infty]<br />

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

 

[Mark44]

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

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.

for #1
the rule states:
"If the expression
<br /> \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}<br />
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."

what about
<br /> [\infty^0],~[\infty - \infty]<br />

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.

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

Sort of.
\lim_{x \rightarrow 0}\left[ 1/x - 1/sin(x)\right] is an indeterminate form of this type.

 

[Slimsta]

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.

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 <br /> \lim_{x \rightarrow 0}\left[ 1/x - 1/sin(x)\right]<br />?

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

 

[Mark44]

#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.

 

[Slimsta]

#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.

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? because 1/infinity?

 

[Mark44]

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.

 

[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?

 

[Mark44]

Sorry about that - was thinking about the next step and got ahead of myself.
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?

 

[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!

 

[Mark44]

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.