Limit using L'Hôpital

May 2010
254
8
Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

\(\displaystyle \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}{x^2}}}\)

I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

\(\displaystyle \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}\)

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}{x^2}}\)

And then:

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}\)

How should I proceed?

I've tried using L'Hôpitals rule, but I didn't get too far.

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}\)

From here it would seem to be that the answer is \(\displaystyle e^{-\infty}\) (which I think it would be equal to zero) but I used derive to make the calculus, and the límit should give \(\displaystyle e^{-1/6}\)

Bye there.
 
Oct 2009
4,261
1,836
Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

\(\displaystyle \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}{x^2}}}\)

I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

\(\displaystyle \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}\)

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}{x^2}}\)

And then:

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}\)

How should I proceed?

I've tried using L'Hôpitals rule, but I didn't get too far.

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}\)

Bye there.


\(\displaystyle \left(\ln\frac{\sin x}{x}\right)^{'}=\frac{x}{\sin x}\frac{x\cos x-\sin x}{x^2}\) \(\displaystyle =\frac{x\cos x-\sin x}{x\sin x}\) , so applying L'Hospital you get:

\(\displaystyle \frac{x\cos x-\sin x}{2x^2\sin x}\) , and L'H once again: \(\displaystyle \frac{-x\sin x}{4x\sin x+2x^2\cos x}\) , and once again: \(\displaystyle \frac{-\sin x-x\cos x}{4\sin x+8x\cos x-2x^2\sin x}\) , and again:

\(\displaystyle \frac{-2\cos x+x\sin x}{12\cos x-12x\sin x-2x^2\cos x}\xrightarrow [x\to o^+]{}-\frac{2}{12}=-\frac{1}{6}\) , so the limit is \(\displaystyle e^{-1/6}\)

Tonio
 
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Prove It

MHF Helper
Aug 2008
12,883
4,999
Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

\(\displaystyle \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}{x^2}}}\)

I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

\(\displaystyle \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}\)

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}{x^2}}\)

And then:

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}\)

How should I proceed?

Bye there.
\(\displaystyle \lim_{x \to 0^{+}}\left[\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x^2}}\right] = \lim_{x \to 0^{+}}e^{\ln{\left[\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x^2}}\right]}}\)

\(\displaystyle = \lim_{x \to 0^{+}}e^{\frac{\ln{\left(\frac{\sin{x}}{x}\right)}}{x^2}}\)

\(\displaystyle = e^{\lim_{x \to 0^{+}}\frac{\ln{\left(\frac{\sin{x}}{x}\right)}}{x^2}}\)

This \(\displaystyle \to \frac{0}{0}\), so you can use L'Hospital's Rule...

\(\displaystyle e^{\lim_{x \to 0^{+}}\frac{\ln{\left(\frac{\sin{x}}{x}\right)}}{x^2}} = e^{\lim_{x \to 0^{+}}\left(\frac{\frac{\frac{x\cos{x} - \sin{x}}{x^2}}{\frac{\sin{x}}{x}}}{2x}\right)}\)

\(\displaystyle = e^{\lim_{x \to 0^{+}}\left(\frac{\frac{x\cos{x} - \sin{x}}{x\sin{x}}}{2x}\right)}\)

\(\displaystyle = e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)}\).

This still \(\displaystyle \to \frac{0}{0}\) so use L'Hospital's Rule again...

\(\displaystyle e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(\frac{\cos{x} - x\sin{x} - \cos{x}}{2x^2\cos{x} + 4x\sin{x}}\right)}\)

\(\displaystyle = e^{\lim_{x \to 0^{+}}\left(-\frac{\sin{x}}{4\sin{x} + 2x\cos{x}}\right)}\)

This still \(\displaystyle \to \frac{0}{0}\), so use L'Hospital's Rule again...

\(\displaystyle e^{\lim_{x \to 0^{+}}\left(-\frac{\sin{x}}{4\sin{x} + 2x\cos{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(-\frac{\cos{x}}{4\cos{x} + 2x\sin{x} + 2\cos{x}}\right)}\)

\(\displaystyle = e^{-\frac{1}{4 + 2}}\)

\(\displaystyle = e^{-\frac{1}{6}}\).



PHEW!

Edit: Fixed a small mistake.
 
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Prove It

MHF Helper
Aug 2008
12,883
4,999
\(\displaystyle \left(\ln\frac{\sin x}{x}\right)^{'}=\frac{x}{\sin x}\frac{x\cos x-\sin x}{x^2}\) \(\displaystyle =\frac{x\cos x-\sin x}{x\sin x}\) , so applying L'Hospital you get:

\(\displaystyle \frac{x\cos x-\sin x}{2x^2\sin x}\) , and L'H once again: \(\displaystyle \frac{-x\sin x}{4x\sin x+2x^2\cos x}\) , and once again: \(\displaystyle \frac{-\sin x-x\cos x}{4\sin x+8x\cos x-2x^2\sin x}\) , and again:

\(\displaystyle \frac{-2\cos x+x\sin x}{12\cos x-12x\sin x-2x^2\cos x}\xrightarrow [x\to o^+]{}-\frac{2}{12}=-\frac{1}{6}\) , so the limit is \(\displaystyle e^{-1/6}\)

Tonio
You've forgotten about the \(\displaystyle 2x\) in the denominator that comes from the power in the logarithm.
 

chiph588@

MHF Hall of Honor
Sep 2008
1,163
429
Champaign, Illinois
In general to get the \(\displaystyle 1^\infty \) indeterminate form into the form \(\displaystyle \frac00 \) do the following:

If \(\displaystyle \lim_{x\to a}f(x) = 1 \) and \(\displaystyle \lim_{x\to a}g(x) = \infty \)

then \(\displaystyle \lim_{x\to a} f(x)^{g(x)} = \exp\left(\lim_{x\to a}\frac{\ln f(x)}{1/g(x)}\right) \).

Then just apply L'Hôpital's rule to get your answer.
 
May 2010
254
8
I wanted to ask him about that Prove it. Thanks. And I know that the result that he get to its good, because its the same that Derive gave it to me.

I have to go now, but I'll read this later.

Bye there, and thanks to both of you.
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

\(\displaystyle \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}{x^2}}}\)

I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

\(\displaystyle \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}\)

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}{x^2}}\)

And then:

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}\)

How should I proceed?

I've tried using L'Hôpitals rule, but I didn't get too far.

\(\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}\)

From here it would seem to be that the answer is \(\displaystyle e^{-\infty}\) (which I think it would be equal to zero) but I used derive to make the calculus, and the límit should give \(\displaystyle e^{-1/6}\)

Bye there.

We have \(\displaystyle \frac{\sin(x)}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - ... \)


\(\displaystyle \lim_{x\to 0} \left( \frac{\sin(x)}{x} \right)^{\frac{1}{x^2}} = \lim_{x\to 0} \left( 1 - \frac{x^2}{6} + \frac{x^4}{120} - ...\right)^{\frac{1}{x^2}}\)

Let \(\displaystyle n = \frac{1}{x^2} \) and we have

\(\displaystyle \lim_{n \to \infty} \left( 1 - \frac{1}{n}( \frac{1}{6} - \frac{1}{120 n} - ...) \right )^n \)

\(\displaystyle = exp \{\ -( \lim_{n \to \infty} \frac{1}{6} - \frac{1}{120 n} - ... ) \}\ \)

\(\displaystyle = e^{-1/6}\)

EDIT: I think it is unnecessary to evaluate only the right hand limit .
 
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simplependulum

MHF Hall of Honor
Jan 2009
715
427
\(\displaystyle
e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(\frac{\cos{x} - x\sin{x} - \cos{x}}{2x^2\cos{x} - 4x\sin{x}}\right)}
\)
Mistake appears in this line

The derivative of \(\displaystyle 2x^2 \sin(x) \) is

\(\displaystyle 4x\sin(x) + 2x^2 \cos(x) \)