Limit Continuity And Differentiability Ques 46
- Let $\alpha \in R$. Prove that a function $f: R \rightarrow R$ is differentiable at $\alpha$ if and only if there is a function $g: R \rightarrow R$ which is continuous at $\alpha$ and satisfies $f(x)-f(\alpha)=g(x)(x-\alpha), \forall x \in R$.
(2001, 5M)
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Solution: Since, $g(x)$ is continuous at $x=\alpha \Rightarrow \lim _{x \rightarrow \alpha} g(x)=g(\alpha)$
and $f(x)-f(\alpha)=g(x)(x-\alpha), \forall x \in R \quad $ [given]
$\Rightarrow \quad \frac{f(x)-f(\alpha)}{(x-\alpha)}=g(x)$
$\Rightarrow \quad \lim _{x \rightarrow \alpha} \frac{f(x)-f(\alpha)}{x-\alpha}=\lim _{x \rightarrow \alpha} g(x)$
$\Rightarrow \quad f^{\prime}(\alpha)=\lim _{x \rightarrow \alpha} g(x) \Rightarrow f^{\prime}(\alpha)=g(\alpha)$
$\Rightarrow \quad f(x)$ is differentiable at $x=\alpha$.
Conversely, suppose $f$ is differentiable at $\alpha$, then $\lim _{x \rightarrow \alpha} \frac{f(x)-f(\alpha)}{x-\alpha}$ exists finitely.
Let $\quad g(x)=\left\{\begin{array}{cc}\frac{f(x)-f(\alpha)}{x-\alpha}, & x \neq \alpha \\ f^{\prime}(\alpha), & x=\alpha\end{array}\right.$
Clearly, $\quad \lim _{x \rightarrow \propto} g(x)=f^{\prime}(\alpha)$
$\Rightarrow \quad g(x)$ is continuous at $x=\alpha$.
Hence, $f(x)$ is differentiable at $x=\alpha$, iff $g(x)$ is continuous at $x=\alpha$.
BETA
