Proof of a limit - Beginner

Question

I am learning about limits and I did some exercises but this one in particular I am not getting.

Let $\mathbb R_{\ne 0} \rightarrow \mathbb {R}, f(x)=\frac{1}{x}.$

Proof that $\lim_{x \to \frac{1}{2}} f(x) = 2$

I can't find $\delta$ that satisfies the properties:

$|f(x)-2|<\varepsilon$ and $0<|x-\frac{1}{2}|<\delta$

Can you help me ?

Answer 1

$0< |x - \frac 12| < \delta \implies$

$-\delta < x -\frac 12 < \delta; x \ne \frac 12$

$\frac 12 -\delta < x < \frac 12 + \delta$. As we can choose $\delta$ to be small will make sure at the end that is is less than $\frac 12$.

$\frac 1{\frac 12 + \delta} < \frac 1 x < \frac 1{\frac 12 - \delta}$

$\frac 2{1 + 2\delta} < \frac 1x < \frac 2{\frac 1- \delta}$

$\frac 2{1 + 2\delta} - 2 < \frac 1x -2 < \frac 2{1-2\delta} -2 $

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

$\frac {-4\delta}{1+2\delta} < f(x) -2 < \frac {4\delta}{1 -2\delta}$

So we want

$-\epsilon \le \frac {-4\delta}{1+2\delta} < f(x) -2 < \frac {4\delta}{1 - 2\delta}< \epsilon$

And therefore $|f(x) - 2| < \epsilon$

But what $\delta$ can we use that will make

$-\epsilon \le \frac {-4\delta}{1+2\delta}$, or in other words $\epsilon \ge \frac {4\delta}{1+2\delta}$, and $\epsilon \ge \frac {4\delta}{1-2\delta}$?

Well, since $\frac {4\delta}{1+2\delta} < \frac {4\delta}{1-2\delta}$ we just need $\epsilon \ge \frac {4\delta}{1-2\delta}$.

So we just want $\delta < \frac 12$ and $4\delta \le \epsilon (1-2\delta)$

$\delta(4+2\epsilon) \le \epsilon$ or

$\delta \le \frac {\epsilon}{4 + 2\epsilon}$.

.....

so for example if $\epsilon = .01$ we can choose $\delta \le \frac {.01}{4.02}$ so lets take $\delta = .002$.

If $|x - .5| < .002$ then $.498 < x < .502$ and

$1.992 < \frac 1{.502} < \frac 1x < \frac 1{.498} < 2.0081$

$-.008 < \frac 1x -2 < .0081$

$|f(x) - 2| < .0081 < .01$.

Answer 2

The main problem is that near $0$, $1/x$ is unbounded. Here is a hint: To fix this, you want to ensure that your $\delta$ is always less than a specified value, so that you can bound the values of $f(x)$. For example, when $|x-1/2| < 1/4$, $4/3 < f(x) < 4$. So once you insist that $\delta \leq 1/4$, this can be approached as a normal $\epsilon$-$\delta$ limit problem.