Evaluate $\lim_{x\to 0^+}\frac{\arccos(x)-\frac{\pi}{2}}{x\ln(x)}$

Question

Evaluate $$\lim_{x\to 0^+}\frac{\arccos(x)-\frac{\pi}{2}}{x \ln x}$$

$$\lim_{x\to 0^+}\frac{\arccos(x)-\frac{\pi}{2}}{x \ln x}=\lim_{x\to 0^+}\frac{\arccos(x)-\arccos(0)}{x-0}\cdot \frac{1}{ \ln x}=-\frac{1}{\sqrt{1-0^2}}\cdot0=0$$

Is it valid?

Answer 1

Yes, because if we know that $\lim_{x\to 0^+}f(x)$ and $\lim_{x\to 0^+}g(x)$ exist, then we have that $\lim_{x\to 0^+}f(x)g(x)$ exists and $\lim_{x\to 0^+}f(x)g(x)=\lim_{x\to 0^+}f(x)\lim_{x\to 0^+}g(x)$.

Answer 2

I think your answer is right.

You used the definition of the derivative.

Also, we can make the following.

$$\lim_{x\rightarrow0^+}\frac{\arccos{x}-\frac{\pi}{2}}{x\ln{x}}=\lim_{x\rightarrow0^+}\frac{-\frac{1}{\sqrt{1-x^2}}}{\ln{x}+1}=0.$$

Answer 3

Yes it's fine!

As an alternative note that:

$$\arccos x=\frac{\pi}{2}-\arcsin x$$

thus

$$\frac{\arccos x-\frac{\pi}{2}}{x\ln x}=-\frac{\arcsin x}{x}\frac{1}{\ln x}\to -1\cdot 0=0 $$