Why $\lim_{x \to 0} \left(\frac{\sin(x)}{x}\right)^\frac{1}{x} = 1$? I understood it's solution using Taylor series for sin x but main concern is that isn't $$\frac{\sin(x)}{x}<1$$ So shouldn't limit of a fraction raised to power of number tending to infinity should tend to $0$?
 A: Not necessarily. The limit of $f(x)^{g(x)}$ for $x\to \infty$ where $\lim_{x\to \infty} f(x)=1$ and $\lim_{x\to \infty} g(x)=\infty$ can vary based on how strong $f(x)$ and $g(x)$ tend to their limits. For example,
$$
{\lim_{n\to \infty} \left(1-\frac{1}{n^2}\right)^n=1,
\\\lim_{n\to \infty}  \left(1-\frac{1}{n}\right)^{an}=\frac{1}{e^a},
\\\lim_{n\to \infty}  \left(1-\frac{1}{n}\right)^{n^2}=0.
}
$$
A: There are competing tendencies in this indeterminate form $A^B$. The power $B$ is growing arbitrarily large while $A$ is increasing to $1$. To see what the limit is takes work.
For an analogy, remember that
$$
\left( 1 + \frac{1}{n} \right)^n
$$
approaches $e$ as $n$ grows.
A: $$L=\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^\frac{1}{x} $$
$$\ln L=\lim_{x \to 0} \frac{\ln \left(\frac{\sin x}{x}\right)}{x} $$
This is of the form $0/0$ so use l'Hopital:
$$\ln L=\lim_{x \to 0} \frac{\ln \sin x - \ln x}{x}=\lim_{x \to 0} \frac{\cos x}{ \sin x} - \frac1{x} =\lim_{x \to 0} \frac{x\cos x - \sin x}{x \sin x}$$
We now use the small angle approximations:
$$\ln L=\lim_{x \to 0} \frac{x(1-\frac12 x^2) - (x-\frac16 x^3)+O(x^5)}{x (x-\frac16 x^3)+O(x^5)}=\lim_{x \to 0} \frac{-\frac13 x^3 +O(x^5)}{x^2+O(x^4) }$$
$$\ln L=\lim_{x \to 0} \frac{-\frac13 x +O(x^3)}{1+O(x^2) }=0$$
And so the limit is $L=1$.
