1. ## Limit using L'Hôpital

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

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

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

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

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

And then:

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

How should I proceed?

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

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

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

Bye there.

2. Originally Posted by Ulysses
Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

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

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

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

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

And then:

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

How should I proceed?

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

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

Bye there.

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

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

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

Tonio

3. Thank you very much Tonio!

4. Originally Posted by Ulysses
Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

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

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

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

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

And then:

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

How should I proceed?

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

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

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

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

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

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

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

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

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

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

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

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

$= e^{-\frac{1}{4 + 2}}$

$= e^{-\frac{1}{6}}$.

PHEW!

Edit: Fixed a small mistake.

5. Originally Posted by tonio
$\left(\ln\frac{\sin x}{x}\right)^{'}=\frac{x}{\sin x}\frac{x\cos x-\sin x}{x^2}$ $=\frac{x\cos x-\sin x}{x\sin x}$ , so applying L'Hospital you get:

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

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

Tonio
You've forgotten about the $2x$ in the denominator that comes from the power in the logarithm.

6. In general to get the $1^\infty$ indeterminate form into the form $\frac00$ do the following:

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

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

7. 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.

8. Originally Posted by Ulysses
Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

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

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

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

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

And then:

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

How should I proceed?

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

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

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

Bye there.

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

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

Let $n = \frac{1}{x^2}$ and we have

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

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

$= e^{-1/6}$

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

9. $
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 $2x^2 \sin(x)$ is

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

10. Originally Posted by simplependulum
Mistake appears in this line

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

$4x\sin(x) + 2x^2 \cos(x)$
Thanks.

11. Originally Posted by Prove It
You've forgotten about the $2x$ in the denominator that comes from the power in the logarithm.

Where exactly? I don't think I'd gottent the correct answer had I forgotten that, but perhaps I did. Where?

Tonio

12. Well, the result that he gets its the right one.

It's great to have so many focus for the same problem. Thanks to all of you.

Originally Posted by simplependulum
We have $\frac{\sin(x)}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - ...$

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

Let $n = \frac{1}{x^2}$ and we have

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

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

$= e^{-1/6}$

EDIT: I think it is unnecessary to evaluate only the right hand limit .
Are you using Taylors polinomial there? I haven't seen it yet, but its good to know it can be solved that way.

Bye there.

EDIT: Now that I've taken a deeper look on tonios reasoning i've seen that he proceeded right and didn't forget a thing. It's just that as he started by reasoning the logaritm part I thought that he forgot the denominator, but now I see that he didn't.