# 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 $\frac{1}{x}$??

• Shouldn't that be $O(x^5)$? Commented Dec 11, 2014 at 22:14
• @ThomasAndrews Technically since it is $O$ and not $\Theta$, the form above is correct even if it is imprecise. The $O(x^2)$ should be $O(x^4)$ or at least $o(x^3)$, though, with the corresponding change to the $O(x)$.
– Ian
Commented Dec 11, 2014 at 22:17

$$\left(\frac{\sin x }{x}\right)^{\frac1x} = e^{\frac 1x \ln{\frac{\sin x}{x}}} = e^{\frac 1x \ln({1 - x^2 / 6 + O(x^4) })} = e^{\frac 1x(-x^2/6 + O(x^4))} = e^{-x/6 + O(x^3)} \to 1$$

• quick question: how is $ln(1-x^2/6+O(x^4)) = (-x^2/6 + O(x^4))$?? Commented Dec 11, 2014 at 22:34
• @rndflas it comes from the taylor series of the logarithm.. or just $\ln ( 1 + x) = x + O(x)$
– Ant
Commented Dec 11, 2014 at 22:36

Compose with $\mathbb{e}^x$ :

$$\left(h(x)\right)^{g(x)} = \mathbb{e}^{g(x)\ln{h(x)}}$$

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

• how is $e = (1+\frac{x^2}{6})$?? Commented Dec 11, 2014 at 22:32
• @rndflas $$\lim_{x \to 0} \left (1 + \frac {x^2}6\right )^{\frac 6{x^2}} = \lim_{u \to 0} \left ( 1 + u\right )^{\frac 1u} = e$$ Commented Dec 11, 2014 at 23:46