MHB How Do We Prove the Limit of 1/x as x Approaches 2?

[Dethrone]

$$\lim_{{x}\to{2}}\frac{1}{x}=\frac{1}{2}$$

Here is what I have so far:

For all $\delta >0$, there exists an $x$ such that $0<|x-2|<\delta $, $|\frac{1}{x}-\frac{1}{2}<\epsilon$

Cut to the chase:
$$\frac{|x-2|}{|2x|}<\epsilon$$

I need to bound $\frac{1}{|2x|}$ somehow, and represent it with $M$ such that
$M|x-2|<M \delta$.

I'm having trouble finding a reasonable bound for $\delta$, I have tried $\delta<1$:

Blah blah blah:

$2<2x<6$ and $|\frac{1}{2x}|<\frac{1}{2}$.

Now we have $\frac{|x-2|}{|2x|}<\frac{|x-2|}{2}<\epsilon$, so $|x-2|<2 \epsilon$. Now let $\delta_{min}=({1, 2\epsilon})$, but I don't think this will work, any ideas? (Wondering)

 

[Euge]

$$\lim_{{x}\to{2}}\frac{1}{x}=\frac{1}{2}$$

Here is what I have so far:

For all $\delta >0$, there exists an $x$ such that $0<|x-2|<\delta $, $|\frac{1}{x}-\frac{1}{2}<\epsilon$

Cut to the chase:
$$\frac{|x-2|}{|2x|}<\epsilon$$

I need to bound $\frac{1}{|2x|}$ somehow, and represent it with $M$ such that
$M|x-2|<M \delta$.

I'm having trouble finding a reasonable bound for $\delta$, I have tried $\delta<1$:

Blah blah blah:

$2<2x<6$ and $|\frac{1}{2x}|<\frac{1}{2}$.

Now we have $\frac{|x-2|}{|2x|}<\frac{|x-2|}{2}<\epsilon$, so $|x-2|<2 \epsilon$. Now let $\delta_{min}=({1, 2\epsilon})$, but I don't think this will work, any ideas? (Wondering)

Hi Rido12,

While you have the right idea in the last part of your post, it looks like you're getting confused with the $\epsilon-\delta$ definition of a limit. To show that $\lim_{x\to 2} \frac{1}{x} = \frac{1}{2}$, we need to prove that for a given $\epsilon > 0$, there corresponds a $\delta > 0$ such that for all $x$, $0 < |x - 2| < \delta$ implies $$|\frac{1}{x} - \frac{1}{2}| < \epsilon.$$

As a preliminary analysis, we attempt to express $|\frac{1}{x} - \frac{1}{2}|$ in terms of $|x - 2|$:

$$|\frac{1}{x} - \frac{1}{2}| = \frac{|x - 2|}{|2x|}$$

Suppose $\delta = 1$. Then $0 < |x - 2| < \delta$ implies $1 < x < 3$ and $x \neq 2$. Thus $|2x| = 2x > 2$ and

$$|\frac{1}{x} - \frac{1}{2}| < \frac{|x - 2|}{2}$$

Note that $|x - 2|/2 < \epsilon$ implies $|x - 2| < 2\epsilon$. So by choosing $\delta$ to be the smaller of the numbers $1$ and $2\epsilon$ (i.e., $\delta = \min(1, 2\epsilon)$), we are guaranteed that $|\frac{1}{x} - \frac{1}{2}| < \epsilon$ whenever $0 < |x - 2| < \delta$.