Proof of a limit using the epsilon-delta definition of a limit

Quote:

Originally Posted by csstudent5000

I know how to do proofs of limits using the epsilon-delta definition of a limit. However I'm struggling with this particular problem:

$\lim_{x \to 1} \frac{2x^2 + 2}{x + 3} = 1$

Is this problem even possible to prove using algebraic manipulation to transform $|\frac{2x^2 + 2}{x + 3} - 1|$ into $|x - 1|$ (and arrive at some expression of delta in terms of epsilon)? Or does it require some other technique which I'm not aware of?

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

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

Now we need an upper bound for $\displaystyle \begin{align*} \left| 2 - \frac{5}{x + 3}\right| \end{align*}$ . Notice that $\displaystyle \begin{align*} \left| 2 - \frac{5}{x + 3}\right| \leq |2| + \left|- \frac{5}{x + 3} \right| = 2 + 5\left|\frac{1}{x + 3}\right| \end{align*}$ by the triangle inequality.

If we restrict $\displaystyle \begin{align*} |x - 1| < 1 \end{align*}$ (say), then we would have

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

which gives

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

So continuing from $\displaystyle \begin{align*} \left| 2 - \frac{5}{x + 3}\right||x - 1| < \epsilon \end{align*}$ we have

$\displaystyle \begin{align*} \frac{11}{3}|x-1| &< \epsilon \\ |x - 1| &< \frac{3\epsilon}{11} \end{align*}$

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