# Thread: Limit & Derivative Sin x..

1. ## Limit & Derivative Sin x..

1.
$\displaystyle \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}$
2.
$\displaystyle \lim_{x\to 0}(\frac{\sin{3x}}{3x})^\frac{1}{x}$

Should I use just L'Hopital rules or Taylor's teorem?

2. Originally Posted by mmr
1.
$\displaystyle \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}$
I assume you know that $\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}=1$

3. Originally Posted by mmr
1.
$\displaystyle \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}$
2.
$\displaystyle \lim_{x\to 0}(\frac{\sin{3x}}{3x})^\frac{1}{x}$

Should I use just L'Hopital rules or Taylor's teorem?
You could take the natural log and use taylors theorem. Or you could just use Maclaurin series $\displaystyle \frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\cdots\sim{1}$

4. I assume you know that $\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}=1$
Yes but what about the $\displaystyle \frac{1}{x}$ ?

Originally Posted by Mathstud28
Just note that $\displaystyle \sin(x)^{\frac{1}{x}}\to{1}$ and so does $\displaystyle \frac{1}{x^{\frac{1}{x}}}$
Or you could take the natural log and use taylors theorem. Or you could just use Maclaurin series $\displaystyle \frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\cdots\sim{1}$
Ok, don't know Maclaurin series. To simplify the question, could I done it just with L'Hopital rules?

5. Originally Posted by mmr
Yes but what about the $\displaystyle \frac{1}{x}$ ?

Ok, don't know Maclaurin series. To simplify the question, could I done it just with L'Hopital rules?
Yes, just take the natural log first.

6. $\displaystyle \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}=e^{\lim_{x\to 0}\frac{(ln{\frac{\sin{x}}{x}})'}{(x)'}}=e^{\lim_{ x\to 0}\frac{x\cos{x}-\sin{x}}{x\sin{x}}}$
Is this good? if yes,what next?

7. Originally Posted by mmr
$\displaystyle \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}=e^{\lim_{x\to 0}\frac{(ln{\frac{\sin{x}}{x}})'}{(x)'}}=e^{\lim_{ x\to 0}\frac{x\cos{x}-\sin{x}}{x\sin{x}}}$
Is this good? if yes,what next?
Consider this, as $\displaystyle x\to{0}$ we have that $\displaystyle \sin(x)\to{x}$

So let us rewrite this limit as follows

\displaystyle \begin{aligned}\lim_{x\to{0}}\frac{x\cos(x)-x}{x\sin(x)}&=\lim_{x\to{0}}\frac{x\cos(x)-x}{x^2}\\ &=\lim_{x\to{0}}\frac{1-\cos(x)}{x} \end{aligned}

Now this is a simple application of L'hopital's,