Find $\lim_{x\to 0^{\pm}}f(x)=\frac{\arcsin(1-\left\{x\right\})\times\arccos(1-\left\{x\right\})}{\sqrt{2\left\{x\right\}}\times(1-\left\{x\right\})}$

Question

If $f(x)=\frac{\arcsin(1-\left\{x\right\})\times\arccos(1-\left\{x\right\})}{\sqrt{2\left\{x\right\}}\times(1-\left\{x\right\})}$
Find $\lim_{x\to 0^+}f(x)$ and $\lim_{x\to 0^-}f(x)$ where $\left\{x\right\} $is a fractional part function.

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As $x\to 0^+,\left\{x\right\}\to 0$, so $\lim_{x\to 0^+}\frac{\arcsin(1-\left\{x\right\})\times\arccos(1-\left\{x\right\})}{\sqrt{2\left\{x\right\}}\times(1-\left\{x\right\})}$ becomes $\frac{0}{0}$.

As $x\to 0^-,\left\{x\right\}\to 1$, so $\lim_{x\to 0^+}\frac{\arcsin(1-\left\{x\right\})\times\arccos(1-\left\{x\right\})}{\sqrt{2\left\{x\right\}}\times(1-\left\{x\right\})}$ becomes $\frac{0}{0}$.

Both the limits are eligible for L Hospital rule, but i do not know how to differentiate the fractional part function. The answers given are $\lim_{x\to 0^+}f(x)=\frac{\pi}{2}$ and $\frac{\pi}{2\sqrt2}$ .
Please help me. Thanks

Answer 1

Since $x=\lfloor x\rfloor +\{x\}$, we have $\{x\}=x-\lfloor x\rfloor $.

For $x\to 0^+$, since $\lfloor x\rfloor=0$, we have $\{x\}=x$.

For $x\to 0^-$, since $\lfloor x\rfloor=-1$, we have $\{x\}=x+1$.