This is a problem from Differential Calculus for Beginners by Joseph Edwards

Question

$$\lim_{x\rightarrow 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$$ I attempted to expand $\sin x$ and am unsure how to proceed. $$1-\frac{x}{3!}\left[x^2-\frac{x^4}{5\times 4} \cdots \right]$$

Can anyone please help me with this Problem?

Answer 1

Write it in e to the power ln format . Then u get 1/x(Ln(sin/x)) then put limit for sin/x. I think u got the idea

Answer 2

Let $f(x) = \lim_{x\to0} (\frac{\sin(x)}{x})^{1/x}$.
A nice property of logarithms is that they can bring an exponent within a function in front. Because of this, we have $$\ln(f(x)) = \frac{\ln(\lim_{x\to0}\frac{\sin(x)}{x})}{x}$$

We know from the squeeze theorem that $$\lim_{x\to0}\frac{\sin(x)}{x} = 1$$ Thus, we have $$\ln(f(x)) = \frac{\ln(1)}{x}$$

$\ln(1) = 0$, so we have $$\ln(f(x)) = \frac{0}{x} = 0$$ Following this, we know that $$\ln(f(x)) = 0$$ so we take both sides to the power of e: $$e^{\ln(f(x))} = e^0$$ $$f(x) = 1$$