# Thread: Need help with l'hopital rule question

1. ## Need help with l'hopital rule question

lim (x goes to 0) ( (sinx) / x ) ^ (1/ (x^2) ) = ?

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

If we let $\displaystyle t=sin(x)$, we get:

$\displaystyle \lim_{t\to 0}\left(\frac{t}{arcsin(t))}\right)^{\frac{1}{arcs in^{2}(t)}}$

$\displaystyle =e^{\left(\lim_{t\to 0}\frac{ln(\frac{t}{arcsin(t)})}{arcsin^{2}(t)}\ri ght)}$

Using L'Hopital and hammering at it, we get it whittled down to:

$\displaystyle e^{\left(\frac{-1}{6}\lim_{t\to 0}\sqrt{1-t^{2}}\right)}$

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

3. Thanks for the hint, but i still cant get it down to -1/6. I keep getting stuck in the loop of infinity - infinity.

4. You have to apply L'Hopital several times.

Eventually, we get:

$\displaystyle e^{\left(\frac{1}{2}\cdot\frac{\lim_{t\to 0}\sqrt{1-t^{2}}}{\lim_{t\to 0} -3\sqrt{1-t^{2}}+\lim_{t\to 0} tsin^{-1}(t)}\right)}$

Now, you can see it?.

I hope you can see that. I don't know how to make it bigger.

5. Originally Posted by galactus
I don't know how to make it bigger.
This can be achieved using \displaystyle :

$\displaystyle \exp{\displaystyle \left(\frac{1}{2}\cdot\frac{\lim_{t\to 0}\sqrt{1-t^{2}}}{\lim_{t\to 0} -3\sqrt{1-t^{2}}+\lim_{t\to 0} tsin^{-1}(t)}\right)}$

or

$\displaystyle \exp{ \left(\frac{1}{2}\cdot\frac{\displaystyle\lim_{t\t o 0}\sqrt{1-t^{2}}}{\displaystyle\lim_{t\to 0} -3\sqrt{1-t^{2}}+\lim_{t\to 0} tsin^{-1}(t)}\right)}$

6. I got the answer now. Thanks for the help!!