# How do you prove by definition that the function f(x)= x^2 sin (1/x) is continuous at x=0?

Dec 30, 2017

Not continuous

#### Explanation:

This function as is, is not continuous at ${x}_{0} = 0$ because it is not defined there.

D_f={x$\in$$\mathbb{R}$:x!=0} $=$ $\mathbb{R}$* $=$ $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

This function would be continuous for example,

$f \left(x\right) = \left\{\begin{matrix}{x}^{2} \sin \left(\frac{1}{x}\right) \text{ & "x!=0 \\ 0" & } x = 0\end{matrix}\right.$

Dec 31, 2017

If you re-define the function as Jim suggested, then it would be continuous at $x = 0$ and this can be proven as show below.

#### Explanation:

We must show that ${\lim}_{x \to 0} f \left(x\right) = f \left(0\right) = 0$.

Method 1 (use the $\frac{\epsilon}{\delta}$ definition of a limit):

Let $\epsilon > 0$ be given (this represents an arbitrarily small distance that we'd like the function outputs to be to the limit 0 if $x$ is sufficiently close to 0).

Choose $\delta = \sqrt{\epsilon} > 0$ (this represents the measure of "sufficiently close" and is chosen this way to make the algebra below work out nicely).

Suppose $| x - 0 | < \delta$ so that $- \sqrt{\epsilon} < x < \sqrt{\epsilon}$. Then $0 \le q {x}^{2} < {\left(\sqrt{\epsilon}\right)}^{2} = \epsilon$ and $| {x}^{2} \sin \left(\frac{1}{x}\right) | < \epsilon$ (when $x \ne 0$), since $- 1 \le \sin \left(\frac{1}{x}\right) \le 1$ for all $x \ne 0$ (also note that $| f \left(0\right) | = 0 < \epsilon$).

But this means that $| f \left(x\right) - 0 | < \epsilon$ for all $x$ such that $| x - 0 | < \delta$ and we have shown that ${\lim}_{x \to 0} f \left(x\right) = f \left(0\right) = 0$, making $f$ continuous at $x = 0$.

Method 2 (use the Squeeze Theorem ):

The facts that $- 1 \le \sin \left(\frac{1}{x}\right) \le 1$ for all $x \ne 0$ and $f \left(0\right) = 0$ imply that $- {x}^{2} \le f \left(x\right) \le {x}^{2}$ for all $x$. Since ${\lim}_{x \to 0} {x}^{2} = 0$ and ${\lim}_{x \to 0} \left(- {x}^{2}\right) = 0$, it follows that ${\lim}_{x \to 0} f \left(x\right) = 0 = f \left(0\right)$ by the Squeeze Theorem.