# Proof of a limit

• Mar 24th 2009, 12:07 AM
redsoxfan325
Proof of a limit
Prove that $\displaystyle \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{10}$ and from this deduce that $\displaystyle \frac{1}{x^2+1}$ is continuous at $\displaystyle x=3$.

Proposed solution:

$\displaystyle \left |\frac{1}{10} - \frac{1}{x^2+1}\right | = \left | \frac{x^2+1}{10x^2+10} - \frac{10}{10x^2+10}\right | = \frac{1}{10}\left |\frac{x^2-9}{x^2+1}\right | = \frac{1}{10}\left | \frac{|x+3||x-3|}{x^2+1}\right |$

If $\displaystyle 0 < |x-3| < 1$, then $\displaystyle |x+3|<7$ and $\displaystyle x^2+1>10$

Thus, $\displaystyle \frac{1}{10}\left | \frac{|x+3||x-3|}{x^2+1}\right | < \frac{1}{10}\left |\frac{7|x-3|}{10} \right | = \frac{7}{100}|x-3|$.

Thus, let $\displaystyle \delta = \min\left\{1, \frac{100\epsilon}{7}\right\}$ therefore when $\displaystyle |x-3|<\delta, \left |\frac{1}{10} - \frac{1}{x^2+1}\right |< \epsilon$.

Since $\displaystyle \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{3^2+1} = \frac{1}{10}$, we can conclude that $\displaystyle \frac{1}{x^2+1}$ is continuous at $\displaystyle x=3$.

Is that correct, and if not, how do I fix it?

Now, what if I need to prove that $\displaystyle \frac{1}{x^2+1}$ is continuous for all x. How would I find an appropriate delta then?
• Mar 24th 2009, 12:49 AM
Jhevon
Quote:

Originally Posted by redsoxfan325
Prove that $\displaystyle \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{10}$ and from this deduce that $\displaystyle \frac{1}{x^2+1}$ is continuous at $\displaystyle x=3$.

Proposed solution:

$\displaystyle \left |\frac{1}{10} - \frac{1}{x^2+1}\right | = \left | \frac{x^2+1}{10x^2+10} - \frac{10}{10x^2+10}\right | = \frac{1}{10}\left |\frac{x^2-9}{x^2+1}\right | = \frac{1}{10}\left | \frac{|x+3||x-3|}{x^2+1}\right |$

If $\displaystyle 0 < |x-3| < 1$, then $\displaystyle |x+3|<7$ and $\displaystyle x^2+1>10$

Thus, $\displaystyle \frac{1}{10}\left | \frac{|x+3||x-3|}{x^2+1}\right | < \frac{1}{10}\left |\frac{7|x-3|}{10} \right | = \frac{7}{100}|x-3|$.

Thus, let $\displaystyle \delta = \min\left\{1, \frac{100\epsilon}{7}\right\}$ therefore when $\displaystyle |x-3|<\delta, \left |\frac{1}{10} - \frac{1}{x^2+1}\right |< \epsilon$.

very nice, but a little snag: $\displaystyle x^2 + 1 > 10$ is not always true with these conditions.

Quote:

Since $\displaystyle \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{3^2+1} = \frac{1}{10}$, we can conclude that $\displaystyle \frac{1}{x^2+1}$ is continuous at $\displaystyle x=3$.

Is that correct, and if not, how do I fix it?
yes, it is fine to show that $\displaystyle \lim_{x \to x_0}f(x) = f(x_0)$, this means $\displaystyle f(x)$ is continuous at $\displaystyle x_0$

Quote:

Now, what if I need to prove that $\displaystyle \frac{1}{x^2+1}$ is continuous for all x. How would I find an appropriate delta then?
you need to show the definition holds for all points in the domain of the function (which happens to be all real numbers). so let $\displaystyle x_0$ be an arbitrary point in the domain. you must show:

For every $\displaystyle \epsilon > 0$, there exists a $\displaystyle \delta > 0$ such that $\displaystyle x,x_0 \in \text{dom}(f)$ and $\displaystyle 0<|x - x_0| < \delta$ implies $\displaystyle |f(x) - f(x_0)|< \epsilon$

to find your delta, you would use similar manipulations to what you have done so far
• Mar 24th 2009, 07:17 AM
redsoxfan325
OK, so I'll change 10 to 5.

Because $\displaystyle 0 < |x-3| < 1$, then either $\displaystyle 0<x-3<1$, in which case 5 still works ($\displaystyle 5<10<x^2+1$), or $\displaystyle 0>x-3>-1$, so $\displaystyle x^2+1>5$.

Thus I'll let $\displaystyle \delta = \min\left\{1, \frac{50\epsilon}{7}\right\}$.
• Mar 24th 2009, 10:07 AM
hatsoff
Quote:

Originally Posted by redsoxfan325
Prove that $\displaystyle \lim_{x\to 3} \frac{1}{x^2+1} = \frac{1}{10}$ and from this deduce that $\displaystyle \frac{1}{x^2+1}$ is continuous at $\displaystyle x=3$.

We need to show that:

$\displaystyle |\frac{1}{x^2+1}-\frac{1}{x_0^2+1}|<\epsilon$

Which simplifies to:

$\displaystyle \frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}<\epsilon$

But we have to get that beginning with delta:

$\displaystyle |x-x_0|<\delta$

So, let's set $\displaystyle \delta\leq\frac{|x_0|}{2}$

Which gives us:

$\displaystyle |x-x_0|<\delta\leq\frac{|x_0|}{2}\Rightarrow|x-x_0|<\frac{|x_0|}{2}$

$\displaystyle \Rightarrow-\frac{|x_0|}{2}<x-x_0<\frac{|x_0|}{2}$

$\displaystyle \Rightarrow-\frac{5|x_0|}{2}\leq-\frac{|x_0|+4x_0}{2}<x+x_0<\frac{|x_0|+4x_0}{2}\le q\frac{5|x_0|}{2}$

$\displaystyle \Rightarrow|x+x_0|<\frac{5|x_0|}{2}$

$\displaystyle \Rightarrow|x-x_0||x+x_0|<\frac{5|x_0|\delta}{2}$

Observe that: $\displaystyle \frac{1}{(x_0^2+1)(x^2+1)}\leq1$

So: $\displaystyle \frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}\leq|x-x_0||x+x_0|$

$\displaystyle \Rightarrow\frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}<\frac{5|x_0|\delta} {2}$

Now we can just choose $\displaystyle \delta=\min\{\frac{x_0}{2},\frac{2\epsilon}{5|x_0| }\}$

$\displaystyle \Rightarrow \delta\leq\frac{2\epsilon}{5|x_0|}\Rightarrow\frac {5|x_0|\delta}{2}\leq\epsilon$

$\displaystyle \Rightarrow\frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}<\frac{5|x_0|\delta} {2}\leq\epsilon$

$\displaystyle \Rightarrow\frac{|x-x_0||x+x_0|}{(x_0^2+1)(x^2+1)}<\epsilon$

So $\displaystyle \lim_{x\to x_0}\frac{1}{x^2+1}=\frac{1}{x_0^2+1}\;\forall\;x\ in\mathbb{R}$

Which means $\displaystyle \frac{1}{x^2+1}$ is continuous for all real $\displaystyle x$.