Question

If f (x) is differentiable at x = c, then write the value of $\underset{x\to c}{\lim }f\left(x\right)$.

Answer 1

Given: $f\left(x\right)$ is differentiable at $x=c$. Then,
$\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ exists finitely.

or, $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=f\text{'}\left(c\right)$.

Consider,
$$\begin{gathered} \underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }\left[\left\{\frac{f\left(x\right)-f\left(c\right)}{x-c}\right\}(x-c)+f\left(c\right)\right] \\ \underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }\left[\left\{\frac{f\left(x\right)-f\left(c\right)}{x-c}\right\}(x-c)\right]+f\left(c\right) \\ \underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }\left\{\frac{f\left(x\right)-f\left(c\right)}{x-c}\right\}\underset{x\to c}{\lim }(x-c)+f\left(c\right) \\ \underset{x\to c}{\lim }f\left(x\right)=f\text{'}\left(c\right)\times 0+f\left(c\right)=f\left(c\right) \end{gathered}$$