Q. 5 TF - Calculus

Question

Interesting trigonometric limits: For each of the functions that follow, use a calculator or other graphing utility to examine the graph of f near $x=0$. Does it appear that f is continuous at $x=0$? Make sure your calculator is set to radian mode  

The function $f$ is continuous at $x=c.$

Step-by-Step Solution

Answer

The proof is $\underset{x\to x_0}{\lim }\left(f\right(x)-f(c\left)\right)=0$

Step 1. Given information

The function $f$ is differentiable at $x=c$ then the function $f$ is continuous at $x=c.$

Step 2. Calculation

A function is said to be continuous over a range if it is graph is a single and unbroken curve.

Formally,

A real valued function $f\left(x\right)$ is said to be continuous at a point $x=x_0$ in domain if

$\underset{x\to x_0}{\lim }f\left(x\right)$ exists and is equal to $f\left(x_0\right)$

If a function $f\left(x\right)$ is continuous at $x=x_0$ then

$\underset{x\to x_0+}{\lim }f\left(x\right)=\underset{x\to x_0-}{\lim }f\left(x\right)=\underset{x\to x_0}{\lim }f\left(x\right)$

Function that are not continuous are said to be discontinuous

Hence proved, 

$$\begin{gathered} \underset{x\to x_0}{\lim }f\left(x\right)=\underset{x\to c}{\lim }f\left(c\right) \\ \underset{x\to x_0}{\lim }f\left(x\right)-\underset{x\to c}{\lim }f\left(c\right)=0 \\ \underset{x\to x_0}{\lim }\left(f\right(x)-f(c\left)\right)=0 \end{gathered}$$