Let f and g be real valued functions defined on interval $(-1,1)$ such that $g''(x)$ is continuous, $g(0)=0, g''(0) \neq 0$, and $f(x)=g(x) \sin x$, $g'(0)=0$.
Statement-1: $\lim\limits_{x \rightarrow 0}\{g(x) \cot x-g(0) ~cosec x \}=f''(0)$
Statement-2: $f'(0)=g(0)$
Answer & explanation
Correct answer: option 2
$\lim\limits_{x \rightarrow 0}\{g(x) \cot x-g(0) ~cosec x\}$
$=\lim\limits_{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x}$ ($\frac{0}{0}$ form)
$=\lim\limits_{x \rightarrow 0} \frac{g'(x) \cos x-g(x) \sin x}{\cos x}$ [Using L' Hospital's Rule]
= 0 [∵ g'(0) = 0 and g(0) = 0]
Now, f(x) = g(x) sin x
$\Rightarrow f'(x)=g'(x) \sin x+g(x) \cos x$ ......(i)
$\Rightarrow f''(x)=g''(x) \sin x+2 g'(x) \cos x-g(x) \sin x$
$\Rightarrow f''(0) = 0$
∴ $\lim\limits_{x \rightarrow 0}\{g(x) \cot x-g(0) ~cosec x\}=f''(0)$ [∵ f''(0) = 0]
So, statement - 1 is true.
From (i), we have f'(0) = g(0)
So, statement - 2 is true. But, statement - 2 is not a correct explanation for statement - 1.