Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0)
eq 0, g^{\prime}(0)=0, g^{\prime \prime}(0)
eq 0$ and $f(x)=g(x) \sin x$. Statement $1 \lim _{x ightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime…

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Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) 
eq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) 
eq 0$ and $f(x)=g(x) \sin x$. Statement $1 \lim _{x ightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$. Statement $2 f^{\prime}(0)=g(0)$.

Aicharya · Accepted Answer

We have, $ \begin{aligned} & \lim _{x ightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x} \ & =\lim _{x ightarrow 0} \frac{g^{\prime}(x) \cos x-g(x) \sin x}{\cos x}=0 \end{aligned} $ Since, $f(x)=g(x) \sin x$ $ \begin{aligned} & \Rightarrow \quad f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x \ & \Rightarrow \quad f^{\prime \prime}(x)=g^{\prime \prime}(x) \sin x+2 g^{\prime}(x) \cos x-g(x) \sin x \ & \Rightarrow \quad f^{\prime \prime}(0)=0 \end{aligned} $ Thus, $\lim _{x ightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=0=f^{\prime \prime}(0)$ $\Rightarrow$ Statement 1 is true. Statement 2. $f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x$ $ \Rightarrow \quad f^{\prime}(0)=g(0) $ Statement 2 is not a correct explanation of Statement 1

Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is…

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