Q. 96 - Calculus

Question

Prove that if a function $f$ is differentiable at $x = c$, then f is continuous at $x = c$.

(a) We are given that f is differentiable at $x = c$. Use the alternative definition of the derivative to write down what that statement means.

(b) We want to show that f is continuous at $x = c$. Use the definition of continuity to show that this statement is equivalent to the statement $\underset{x\to c}{\lim }( f (x) - f (c\left)\right) = 0$.

(c) Now use part (a) to show that $\underset{x\to c}{\lim }( f (x) - f (c\left)\right) = 0$.

(Hint: Multiply ( $f\left(x\right)-f\left(c\right) by \frac{x-c }{x-c}$and use the product rule for limits.)

Step-by-Step Solution

Answer

Ans:

Part (a).$\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=f\left(c\right) it is continuous at x=c$

Part (b).$\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }f\left(x\right) =f\left(c\right)$

Part (c).$\underset{x\to c}{\lim }\left[f\right(x)-f(c\left)\right]=0$

Step 1. Given information:

function $f$ is differentiable at $x=c$

Step 2. Solving Part (a):

Since  $f$ is differentiable at $x=c$

$$\begin{gathered} \underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=f\text{'}\left(c\right) \\ \frac{\underset{x\to c}{\lim }\left(f\left(x\right)-f\left(c\right)\right)}{\underset{x\to c}{\lim }\left(x-c\right)}=f\text{'}\left(c\right) u\sin g division rule \\ \underset{x\to c}{\lim }\left(f\left(x\right)-f\left(c\right)\right)=f\text{'}\left(c\right) \left[\underset{x\to c}{\lim }\left(x-c\right)\right] \\ \underset{x\to c}{\lim }f\left(x\right)-\underset{x\to c}{\lim }f\left(c\right) =f\text{'}\left(c\right)·0 \\ \underset{x\to c}{\lim }f\left(x\right)=f\left(c\right) \\ Hence it is continuous at x=c \end{gathered}$$

Step 3. Solving Part (b):

A function is said to be continuous at $\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }f\left(x\right) =f\left(c\right)$

The limit of the function exists at $x=c$

$$\begin{gathered} \underset{x\to c}{\lim }f\left(x\right)=f\left(c\right) \\ \underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }f\left(c\right) \\ \underset{x\to c}{\lim }f\left(x\right)-\underset{x\to c}{\lim }f\left(c\right)=0 \\ \underset{x\to c}{\lim }\left[f\left(x\right)-f\left(c\right)\right]=0 \\ Hence both statements are equivalent \end{gathered}$$

Step 4. Solving Part (c):

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