Continuity of $ \frac{1}{x^2} $ at $ x=1 $

Question

I have got $ \mid \frac{1}{x^2} - 1 \mid = { \mid x -1 \mid} \frac{1+x}{x^2} $ which would suggest taking $ \delta > \frac{1+x}{x^2}\epsilon $, but $ \frac{1+x}{x^2} $ is unbounded, so I don't understand how this would be done.

Any help is appreciated

Answer 1

but $ \frac{1+x}{x^2} $ is unbounded

It is unbounded only if when $x$ approaches $0$, but we want to prove the continuity at $x=1$. Therefore, we can set a bound $\delta_1=1/2,$ when $|x-1|<\delta_1\Leftrightarrow\frac{1}2<x<\frac{3}2$, in this case $\frac{1+x}{x^2}< 6$ is bounded. So you can set $\delta_2=\frac{\epsilon}6,$ Can you proceed from here?

Answer 2

With arbitrary $~\epsilon > 0,~$ you want

$$-\epsilon < \frac{1}{x^2} - 1 < \epsilon \iff $$

$$1 - \epsilon < \frac{1}{x^2} < 1 + \epsilon. \tag1 $$

So, assume that $~-\delta < x - 1 < \delta.$

This implies that

$$(1 - \delta) < x < (1 + \delta) \implies $$

$$\frac{1}{1 + \delta} < \frac{1}{x} < \frac{1}{1 - \delta} \implies$$

$$\frac{1}{(1 + \delta)^2} < \frac{1}{x^2} < \frac{1}{(1 - \delta)^2}. \tag2$$

Now compare (1) and (2) above.

Suppose that you could establish a relationship between $~\epsilon~$ and $~\delta~$ so that:

• $(1-\epsilon) < \dfrac{1}{(1 + \delta)^2}.$

• $\dfrac{1}{(1 - \delta)^2} < (1 + \epsilon).$

Then, given (1) and (2) above, $~\color{red}{\text{you are done}}.$

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In order to facilitate a linear $~\epsilon ~: ~\delta ~$ relationship, arbitrarily impose the constraint that

$\delta \leq \dfrac{1}{10} \implies $

• $\delta^2 < \delta. $

• $1 - 2\delta > \dfrac{1}{2}.$

So,

$$(1 + \delta)^2 = 1 + 2\delta + \delta^2 < 1 + 3\delta \implies $$

$$\frac{1}{1 + 3\delta} < \frac{1}{(1 + \delta)^2}. $$

Further,

$$\frac{1}{1 + 3\delta} = 1 - \frac{3\delta}{1 + 3\delta} > 1 - 3\delta.$$

Therefore,

$$1 - 3\delta < \frac{1}{1 + 3\delta} < \frac{1}{(1 + \delta)^2}. \tag3 $$

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Similarly,

$$(1 - \delta)^2 = 1 - 2\delta + \delta^2 > 1 - 2\delta \implies $$

$$\frac{1}{(1 - \delta)^2} < \frac{1}{1 - 2\delta}. $$

Further,

$$\frac{1}{1 - 2\delta} = 1 + \frac{2\delta}{1 - 2\delta} < 1 + \frac{2\delta}{1/2} = 1 + 4\delta.$$

Therefore,

$$\frac{1}{(1 - \delta)^2} < \frac{1}{1 - 2\delta} < 1 + 4\delta. \tag4 $$

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So, given (3) and (4) above, the entire problem reduces to $~\color{red}{\text{forcing}}$

• $\delta \leq \dfrac{1}{10}.$

• $1 - \epsilon < 1 - 3\delta.$

• $1 + 4\delta < 1 + \epsilon.$

So, the following constraint works:

$$\delta = \min\left[ ~\frac{1}{10}, ~\frac{\epsilon}{5} ~\right].$$