Problem :
If $f\left( x \right) =\sin\log _e\left( \frac{\sqrt{4-x^2}}{1-x} \right) $ then find the range of this function.
My approach :
$\frac{\sqrt{4-x^2}}{1-x} >0 \Rightarrow 1-x >0 $ also $4-x^2 >0$
$\Rightarrow x \in (-2,1)$ Domain of f(x) is (-2,1)
Now how to find the range of this function please suggest on this .. thanks..
Best Answer
Note that
$$\lim_{x\to -2^+}\frac{\sqrt{4-x^2}}{1-x}=0\;\;,\;\;\lim_{x\to 1^-}\frac{\sqrt{4-x^2}}{1-x}=\infty$$
and thus
$$\left\{\alpha\in\Bbb R\;;\;\alpha=\log\frac{\sqrt{4-x^2}}{1-x}\;,\;\;x\in(-2,1)\right\}=\Bbb R$$
and from here
$$\text{Im}\left(\sin\log\frac{\sqrt{4-x^2}}{1-x}\right)=[-1,1]$$