Question

If $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=10\text{and}\underset{x\to 0}{lim}\frac{f\left(\sin x\right)}{g\left(x\right){\cos }^{2}x}=5$ with $g\left(0\right)=0$, then the value of $\underset{x\to 0}{lim}\frac{g(1-\cos x)}{x^2}$is

Answer 1

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=10$
So, $f\left(0\right)$ must be $0$.

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}\text{(Indeterminate form)}$
Using L' Hospitals' rule, we get
$\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{1}=10$
$\Rightarrow f^{\prime }\left(0\right)=10...\left(1\right)$

Now, $\underset{x\to 0}{lim}\frac{f\left(\sin x\right)}{g\left(x\right){\cos }^{2}x}\to \text{Indeterminate form}$
Using L' Hospitals' rule, we get
$\underset{x\to 0}{lim}\frac{f^{\prime }\left(\sin x\right)\times \cos x}{g^{\prime }\left(x\right){\cos }^{2}x+g\left(x\right)\times 2\cos x\times (-\sin x)}=5$

$\Rightarrow \frac{f^{\prime }\left(0\right)}{g^{\prime }\left(0\right)}=5\Rightarrow g^{\prime }\left(0\right)=\frac{10}{5}=2...\left(2\right)$
$\underset{x\to 0}{lim}\frac{g(1-\cos x)}{x^2}\to \text{Indeterminate form}$

$=\underset{x\to 0}{lim}\frac{g^{\prime }(1-\cos x)\times \sin x}{2x}=\frac{g^{\prime }\left(0\right)}{2}=\frac{2}{2}=1$