Science:Math Exam Resources/Courses/MATH100/December 2014/Question 10/Solution 1

In order to be continuous at $x=0$ we need $\underset{x\to 0^{-}}{\lim }f(x)=f(0)=\underset{x\to 0^{+}}{\lim }f(x)$

• $f(0)=0$
• $\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{-}}{\lim }x=0$

So it remains to determine which $a$ gives $\underset{x\to 0^{+}}{\lim }{x}^a\sin \left(\frac{1}{x}\right)=0$

• If $a=0$ then $\underset{x\to 0^{+}}{\lim }{x}^a\sin \left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{\lim }\sin \left(\frac{1}{x}\right)$ which simply oscillates and so does not exist.
• If $a<0$ then $x^a$ diverges to $+\mathrm{\infty }$ as $x\to 0^{+}$ and so $f(x)$ does not converge (it gets larger and larger and oscillates wildly).
• Finally if $a>0$ then we can use the Squeeze Theorem. Since $-1\le \sin \left(\frac{1}{x}\right)\le 1$ it holds that $-x^a\le f(x)\le x^a$ And as $x\to 0^{+}$ we know that $\underset{x\to 0+}{\lim }-x^a=\underset{x\to 0+}{\lim }{x}^a=0$

So by the Squeeze Theorem, $f(x)\to 0$ as $x\to 0^{+}$. Thus $f(x)$ is continuous at $x=0$ when $a>0$.