# Two tricky L'Hopital limits

• Jan 11th 2011, 04:40 PM
cubejunkies
Two tricky L'Hopital limits
These are all L'Hopital's limits, so we must use a method of taking natural logartihms, taking derivatives of the top and bottom of a fraction, etc.

First problem:

lim as x approaches 0 of (e^x + x)^(1/x)
attempted solution:
ln (e^x + x)^(1/x)
(1/x)(ln (e^x + x))
(1/x)ln (e^x) + (1/x)(ln x)
1 + (1/x)(ln x)
1 + (1/x)
???

before i continue, i must apologize because I dont know how to make my posts be in a more nice format where the notation is all correct and what not. my apologies.

Second problem:

lim x->∞ ((ln x)^(1/x))

y= (ln x)^(1/x)
ln y = (ln(lnx))/x
= 1/(x(ln x))
= x/ (x + ln x)
= 1/(1/x)
= x
and that definitely cant be right

I also should explain a bit. I was taught to set the function that the limit is being taken of equal to y, taking the ln of both sides, bringing the exponent down as a coefficient, and then in these two cases you arrive upon a fraction, so you derive the numerator and denominator of the fraction until the limit is able to be evaluated, the i can setup the limit again as the ln of the original limit being equal to the value obtained in the previous step, and then raising each side to the power of e and getting the correct answer. This is what I did for the other problems in the set and I got all the right answers. I'm just getting stuck on these ones :/

Thank you!
Anthony
• Jan 11th 2011, 05:04 PM
skeeter
Quote:

lim as x approaches 0 of (e^x + x)^(1/x)
attempted solution:
ln (e^x + x)^(1/x)
(1/x)(ln (e^x + x))
(1/x)ln (e^x) + (1/x)(ln x) ... do not use the product rule. Using L'Hopital, you take the derivative of ln(e^x + x) in the numerator over the derivative of x in the denominator
let $\displaystyle \displaystyle y = (e^x+x)^{\frac{1}{x}}$

$\displaystyle \displaystyle \ln{y} = \frac{\ln (e^x + x)}{x}$

$\displaystyle \displaystyle \lim_{x \to 0} \frac{\ln (e^x + x)}{x}$

L'Hopital (derivative of the numerator over the derivative of the denominator) ...

$\displaystyle \displaystyle \lim_{x \to 0} \frac{e^x + 1}{e^x+x} = 2$

$\displaystyle \ln{y} \to 2 \, \implies \, y \to e^2$
• Jan 11th 2011, 05:10 PM
(1).

$\displaystyle \displaystyle\ L=\lim_{x \to 0}\left[e^x+x\right]^{\frac{1}{x}}$

$\displaystyle \displaystyle\ ln(L)=\lim_{x \to 0}ln\left[e^x+x\right]^{\frac{1}{x}}=\frac{ln\left[e^x+x\right]}{x}$

Using L'Hopital's Rule

$\displaystyle \displaystyle\ ln(L)=\lim_{x \to 0}\frac{e^x+1}{e^x+x}=2$

$\displaystyle L=e^2$
• Jan 11th 2011, 06:14 PM
cubejunkies
thank you both so much! now what about the other one? i guess i realize that the first one on my end was mostly stupid mistakes... but the second problem, me and my friend have been stuck for 30 minutes trying to figure it out without success
• Jan 11th 2011, 06:44 PM
tonio
Quote:

Originally Posted by cubejunkies
thank you both so much! now what about the other one? i guess i realize that the first one on my end was mostly stupid mistakes... but the second problem, me and my friend have been stuck for 30 minutes trying to figure it out without success

$\displaystyle \displaystyle{\left(\ln x\right)^{1/x}=e^{\frac{1}{x}\ln\ln x}}$ , and now apply L'Hospital to the power

of the exponential function and use its continuity.

Tonio
• Jan 11th 2011, 06:53 PM
cubejunkies
i dont understand what you mean exactly :/
• Jan 12th 2011, 04:34 AM
tonio
Quote:

Originally Posted by cubejunkies
i dont understand what you mean exactly :/

Well, that's a problem since you should if you're studying this stuff, otherwise it looks like you want people to

completely solve your problems. Take a peek at your text book and watch carefully how the first problem was solved by others.

Tonio
• Jan 12th 2011, 04:46 AM
cubejunkies
I don't appreciate that comment. I just dont understand the thought process involved of setting the original function equal to one of the later algebraic derivations of the function that I had, and I don't recognize how to apply L'Hopital to that equation as you wrote it. A more specific post by me would have asked for clarification more explicitly, therefore I can understand where youre coming from.

Sorry
Anthony
• Jan 12th 2011, 04:54 AM
Soroban
Hello, cubejunkies!

Pay no attention to the man behind the curtain.
Your approach will work . . .

Quote:

$\displaystyle \displaystyle \lim_{x\to\infty}(\ln x)^{\frac{1}{x}}$

We have: .$\displaystyle y \;=\;(\ln x)^{\frac{1}{x}}$

Take logs: .$\displaystyle \ln y \;=\;\ln\left[(\ln x)^{\frac{1}{x}}\right] \;=\;\frac{1}{x}\ln(\ln x) \;=\;\dfrac{\ln(\ln x)}{x}$

Apply L'Hopital: .$\displaystyle \ln y \;=\;\dfrac{\frac{1}{\ln x}\cdot\frac{1}{x}}{1} \;=\;\dfrac{1}{x\ln x}$

$\displaystyle \displaystyle\text{Hence: }\;\lim_{x\to\infty} \ln y \;=\;\lim_{x\to\infty}\frac{1}{x\ln x} \;=\;0$

Therefore: .$\displaystyle \displaystyle \lim_{x\to\inty}y \;=\;e^0 \;=\;1$

• Jan 12th 2011, 05:02 AM
cubejunkies
ohhhhhhhhhhh I see now, I guess I had taken the derivatives too far, because as you can see one of my preliminary steps had 1/(x(lnx)) but I hadn't realized that this could be evaluated as 0. that makes so much more sense now, thank you very much! :)

Anthony
• Jan 12th 2011, 07:34 AM
Quote:

Originally Posted by cubejunkies
These are all L'Hopital's limits, so we must use a method of taking natural logartihms, taking derivatives of the top and bottom of a fraction, etc.

First problem:

lim as x approaches 0 of (e^x + x)^(1/x)
attempted solution:
ln (e^x + x)^(1/x)
(1/x)(ln (e^x + x))
(1/x)ln (e^x) + (1/x)(ln x)

ln(ab)=lna+lnb

$\displaystyle lna+lnb\ \ne\ ln(a+b)$

1 + (1/x)(ln x)
1 + (1/x)
???

Thank you!
Anthony

You violated a log law for the first one.
• Jan 12th 2011, 12:12 PM
Quote:

Originally Posted by cubejunkies
ohhhhhhhhhhh I see now, I guess I had taken the derivatives too far, because as you can see one of my preliminary steps had 1/(x(lnx)) but I hadn't realized that this could be evaluated as 0. that makes so much more sense now, thank you very much! :)

Anthony

$\displaystyle \displaystyle\lim_{x \to \infty}\frac{1}{xlnx}=\lim_{x \to \infty}\frac{\left[\frac{1}{x}\right]}{lnx}$

The numerator is decreasing to 0, while the denominator is increasing,
hence this limit approaches 0.

$\displaystyle \displaysystyle\frac{1}{xlnx}\rightarrow\frac{x}{x +lnx}$