# Thread: Proof of a limit

1. ## Proof of a limit

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

Proposed solution:

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

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

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

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

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

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

Now, what if I need to prove that $\frac{1}{x^2+1}$ is continuous for all x. How would I find an appropriate delta then?

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

Proposed solution:

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

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

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

Thus, let $\delta = \min\left\{1, \frac{100\epsilon}{7}\right\}$ therefore when $|x-3|<\delta, \left |\frac{1}{10} - \frac{1}{x^2+1}\right |< \epsilon$.
very nice, but a little snag: $x^2 + 1 > 10$ is not always true with these conditions.

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

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

Now, what if I need to prove that $\frac{1}{x^2+1}$ 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 $x_0$ be an arbitrary point in the domain. you must show:

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

to find your delta, you would use similar manipulations to what you have done so far

3. OK, so I'll change 10 to 5.

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

Thus I'll let $\delta = \min\left\{1, \frac{50\epsilon}{7}\right\}$.

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

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

Which simplifies to:

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

But we have to get that beginning with delta:

$|x-x_0|<\delta$

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

Which gives us:

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

$\Rightarrow-\frac{|x_0|}{2}

$\Rightarrow-\frac{5|x_0|}{2}\leq-\frac{|x_0|+4x_0}{2}

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

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

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

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

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

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

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

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

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

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

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