Proving the Continuity of a Function

**Requnix** · September 30th 2012, 03:45 AM

Hi

I am trying to proove that (2x^2 + 2) / (x +3) is continous at lim x -> 1 => f(x) = 1

This is what I've thought so far

|f(x) - 1 |< Epsilon
|(2x^2 -x -1) / (x +3)| < Epsilon
|(x - 1)(x + 2) / (x + 3)| < Epsilon

Now I fail to prove that for some K Epsilon = Delta => |x -1| < Delta, thus proves Continouty.
Please help me, I been trying for several hours... Going to try another problem to cool down.

Thanks!

**Prove It** · September 30th 2012, 04:07 AM

Originally Posted by Requnix

Hi

I am trying to proove that (2x^2 + 2) / (x +3) is continous at lim x -> 1 => f(x) = 1

This is what I've thought so far

|f(x) - 1 |< Epsilon
|(2x^2 -x -1) / (x +3)| < Epsilon
|(x - 1)(x + 2) / (x + 3)| < Epsilon

Now I fail to prove that for some K Epsilon = Delta => |x -1| < Delta, thus proves Continouty.
Please help me, I been trying for several hours... Going to try another problem to cool down.

Thanks!

For a function to be continuous at a point, it has to be defined at that point, and the limit of the function approaching that point is equal to the value of the function at that point.

So to show $\displaystyle \begin{align*} \frac{2x^2 + 2}{x + 3} \end{align*}$ is continuous at $\displaystyle \begin{align*} x = 1 \end{align*}$ , you first need to show that the function is defined at $\displaystyle \begin{align*} x = 1 \end{align*}$ (easy, $\displaystyle \begin{align*} f(1) = \frac{2(1)^2 + 2}{1 + 3} = 1 \end{align*}$ ), and then you need to show $\displaystyle \begin{align*} \lim_{x \to 1}\frac{2x^2 + 2}{x + 3} = 1 \end{align*}$ .

To do this, we need to prove $\displaystyle \begin{align*} 0 < |x - 1| < \delta \implies \left|\frac{2x^2 + 2}{x + 3} - 1 \right| < \epsilon \end{align*}$ . Working on the second inequality we have

$\displaystyle \begin{align*} \left|\frac{2x^2 + 2}{x + 3} - 1 \right| &< \epsilon \\ \left|\frac{2x^2 + 2}{x + 3} - \frac{x + 3}{x + 3} \right| &< \epsilon \\ \left|\frac{2x^2 - x - 1}{x + 3} \right| &< \epsilon \\ \left| \frac{2x^2 - 2x + x - 1}{x + 3} \right| &< \epsilon \\ \left| \frac{2x(x - 1) + 1(x - 1)}{x + 3} \right| &< \epsilon \\ \left| \frac{(x - 1)(2x + 1)}{x + 3} \right| &< \epsilon \\ |x - 1| \left| \frac{2x + 1}{x + 3} \right| &< \epsilon \\ |x - 1| \left| \frac{2x + 6 - 5}{x + 3} \right| &< \epsilon \\ |x - 1| \left| 2 - \frac{5}{x + 3} \right| &< \epsilon \end{align*}$

Now if we restrict $\displaystyle \begin{align*} |x - 1| < 1 \end{align*}$ (say), since we have to make the distance from x small anyway, this gives

$\displaystyle \begin{align*} |x - 1| &< 1 \\ -1 < x - 1 &< 1 \\ 0 < x &< 2 \\ 3 < x + 3 &< 5 \end{align*}$

We need an upper bound for $\displaystyle \begin{align*} \left| 2 - \frac{5}{x + 3} \right| \end{align*}$ . Note that

$\displaystyle \begin{align*} \left|2 - \frac{5}{x + 3} \right| &\leq |2| + \left|- \frac{5}{x + 3} \right| \textrm{ by the Triangle Inequality} \\ &= 2 + \frac{5}{|x + 3|} \\ &< 2 + \frac{5}{3} \textrm{ since we restricted } |x - 1| < 1 \implies x + 3 > 3 \\ &= \frac{11}{3} \end{align*}$

Therefore, as long as we have $\displaystyle \begin{align*} |x - 1| < 1 \end{align*}$ , we have

$\displaystyle \begin{align*} |x - 1|\left| 2 - \frac{5}{x + 3} \right| &< \epsilon \\ \frac{11}{3} |x - 1| &< \epsilon \\ |x - 1| &< \frac{3\epsilon}{11} \end{align*}$

So if we let $\displaystyle \begin{align*} \delta = \min\left\{ 1, \frac{3\epsilon}{11} \right\} \end{align*}$ and reverse the process, you will have your proof

**Plato** · September 30th 2012, 02:52 PM

Originally Posted by Requnix

I am trying to proove that (2x^2 + 2) / (x +3) is continous at lim x -> 1 => f(x) = 1
This is what I've thought so far
|f(x) - 1 |< Epsilon
|(2x^2 -x -1) / (x +3)| < Epsilon
|(x - 1)(2x + 2) / (x + 3)| < Epsilon

Once we get the that final line notice that:
$\left| {\frac{{(x - 1)(2x + 1)}}{{x + 3}}} \right| = \left| {\frac{{2x + 1}}{{x + 3}}} \right|\left| {x - 1} \right|$

That is important because we control the value $|x-1|$ .

Therefore we need to bound $\left| {\frac{{2x + 1}}{{x + 3}}} \right|$ .

So start with $|x-1|<1$ from which it follows that
$1<2x+1<5~\&~3<x+3<5$ .

Now we know that $\left| {\frac{{x + 2}}{{x + 3}}} \right|<\frac{5}{3}~.$

Pick $\delta = \min \left\{ {1,\frac{{3\varepsilon }}{5}} \right\}~.$

**hollywood** · September 30th 2012, 05:50 PM

FYI, there's an excellent paper epsilondelta.pdf in the thread "For those that never learnt well the epsilon-delta proofs" which is one of the four "sticky" threads on this forum. Example 3 is almost identical to this one.

- Hollywood

**Requnix** · October 1st 2012, 01:25 AM

I've made an error, however your way of dealing with the |(x - 1) (x + 2) / (x + 3)| < Epsilon, has been most helpfull and enlightening. Thanks again.
2x^2 - x -1 => (1 (+-) Sqrt(1 + 8)) / 4 Giving x = 1 V -(1/2). But whats most important is that (x -1) is still there

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