Q. 4 - 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 

$f\left(x\right)=\left\{\begin{array}{r}{x}^2\sin \left(\frac{1}{x}\right),\text{ if }x\ne 0\\ 0,\text{ if }x=0\end{array}\right.$

Step-by-Step Solution

Answer

Function is continuous at the given point.

Step 1. Given information

We have to explain that f is continuous at x=0

Step 2. Explanation

Put x=0

We get f(0)=0

$\begin{array}{r}\underset{x\to a^{-}}{lim} f\left(x\right)=f\left(a\right)\\ \underset{x\to a^{+}}{lim} f\left(x\right)=f\left(a\right)\end{array}$

Left continuity

$\underset{x\to a^{-}}{lim} f\left(x\right)=f\left(a\right)$

Right continuity 

$\begin{array}{r}\underset{x\to a^{+}}{lim} f\left(x\right)=f\left(a\right)\\ \underset{x\to 0^{-}}{lim} f\left(x\right)=\underset{x\to 0^{-}}{lim} x^2\sin \left(\frac{1}{x}\right)\end{array}$

Step 3. Solving for right continuity

Dividing and multiplying by $\frac{1}{x}$

$\begin{array}{r}\underset{x\to 0^{-}}{lim} f\left(x\right)=\underset{x\to 0^{-}}{lim} \frac{x^2\sin \left(\frac{1}{x}\right)}{\left(\frac{1}{x}\right)}\\ \underset{x\to 0^{-}}{lim} f\left(x\right)=\underset{x\to 0^{-}}{lim} 0\\ \underset{x\to 0^{-}}{lim} f\left(x\right)=0\\ \underset{x\to 0^{+}}{lim} f\left(x\right)=\underset{x\to 0^{+}}{lim} x^2\sin \left(\frac{1}{x}\right)\end{array}$

Dividing and multiplying by $\frac{1}{x}$

$\begin{array}{r}\underset{x\to 0^{+}}{lim} f\left(x\right)=\underset{x\to 0^{+}}{lim} \frac{x^2\sin \left(\frac{1}{x}\right)}{\left(\frac{1}{x}\right)}\\ \underset{x\to 0^{+}}{lim} f\left(x\right)=\underset{x\to 0^{+}}{lim} 0\\ \underset{x\to 0^{+}}{lim} f\left(x\right)=0\\ \underset{x\to a^{-}}{lim} f\left(x\right)=\underset{x\to a^{+}}{lim} f\left(x\right)=f\left(a\right)\end{array}$