Limit of function using Taylor's Formula To find:
$$\lim_{x \to 0} \left(\frac{ \sin(x)}{x}\right)^\frac{1}{x}$$
by using Taylor's formula.
So I used the Taylor's formula for $\sin(x)$ and got::
$\sin(x) = x - \frac{x^3}{6} + O(x^4)$
And then my function becomes:
$$\lim_{x \to 0} \left(\frac{ x - \frac{x^3}{6} + O(x^2)}{x}\right)^\frac{1}{x} = \lim_{x \to 0} \left(1 + \frac{x^2}{6} + O(x)\right)^\frac{1}{x}$$
After that I lost. How am i suppose to get rid of the power of $\frac{1}{x}$??

$$
\lim_{x \to 0} \left( 1 + \frac {x^2}6\right )^{\frac 1x} = \lim_{x \to 0} \left[\left ( 1 + \frac {x^2}6\right )^{\frac 6{x^2}} \right ]^{\frac x6} = \lim_{x \to 0}e^{\frac x6} = 1
$$
