# Find the limit of $\lim_{x \to 0^+} (\sin x)^x$ using standard limits.

Question:

Calculate the limit of the following expression using $a^b = e^{b \ln a}$

$$\lim_{x \to 0^+} (\sin x)^x$$

Attempt at solution:

$$\lim_{x \to 0^+} e^{x\ln (\sin x)}$$

I need to rewrite the expression $\ln (\sin x)$ since it is indeterminate but I am not sure how. I should not use L' Hospitals rule, but use rewrite the limit using known limits.

• Take the logarithm to get $x \log(sin(x))\approx x\log x$ and apply L'Hospital to $\frac{\log x}{ 1/x}$. Commented Mar 8, 2016 at 10:22
• Thank you for your comment, but I should not use L'Hospitals rule for this question. Commented Mar 8, 2016 at 10:23
• Write $\sin x = x\cdot \frac{\sin x}{x}$. The limit of $x \ln x$ may be considered known. Commented Mar 8, 2016 at 10:24
• Thank you @DanielFischer that makes sense. If you add that as an answer I can accept it. Commented Mar 8, 2016 at 10:28

Consider $$A= \sin^x (x)$$ $$\log(A)=x \log(\sin (x))$$ Now, Taylor series around $x=0$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)=x\big(1-\frac{x^2}{6}\big)+\cdots$$ $$\log(\sin (x))\approx \log(x)+\log(1-\frac{x^2}{6})\approx \log(x)-\frac{x^2}{6}$$ So, $$\log(A)\approx x \log(x)-\frac{x^3}{6}$$ which shows the limit and also how it is approached.
For sure, the problem can simplify since $0^0=1$
$$\sin(x)\sim_0 x\implies \ln(\sin(x))\sim_0 \ln(x)\implies x\ln(\sin(x))\sim_0 x\ln(x)\underset{x\to 0}{\longrightarrow }0$$