Prove that the lim f(x) = 1 using the $\epsilon-\delta$ definition

Question

Show directly (from the $\epsilon-\delta$ definition ) that

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

Attempt

Using $$ \left\lvert \frac{x^{2}-3}{x-1} - 1 \right\rvert = \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert = \left\lvert \frac{(x-2)(x+1)}{x-1} \right\rvert $$

and $|x-2|\leq |x - 1 - 1 | \leq |x-1|+|-1| = |x-1|+1 \leq \delta + 1 = 2$ (if $\delta \leq$ 1)

and $|x-1| = |x+1-2| \geq |-2| - |x+1| > 2 - \delta \geq 1$ (if |x+1| < $\delta$)

hence $$ \left\lvert \frac{x^{2}-3}{x-1} - 1 \right\rvert = \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert = \left\lvert \frac{(x-2)(x+1)}{x-1} \right\rvert < 2 \delta \leq \epsilon $$

so $\delta = \frac{\epsilon}{2}$

and picking $\delta$ = min(1,$\frac{\epsilon}{2}$)

hence $$ \forall \epsilon>0 \exists \delta = \min(1,\frac{\epsilon}{2}) > 0 : \forall 0 < \lvert x--1\rvert < \delta = \min(1,\frac{\epsilon}{2}) \implies \left\lvert\frac{x^{2}-3}{x-1} - 1 \right\rvert = \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert = \left\lvert \frac{(x-2)(x+1)}{x-1} \right\rvert < 2 \delta \leq \epsilon$$

hence

$$ \lim_{x\to-1}(\frac{x^{2}-3}{x-1}) = 1$$

Answer 1

I find it easier to make simplifying assumptions first. You have an expression $| { (x-2)(x+1) \over x-1 } |$ which you would like to bound near $x=-1$. Start by bounding $\delta$ by, say, $1$ (that is $|x+1| < 1$), in which case you have $|x-1| > 1$, and $|x-2| < 4$. Then you have $| { (x-2)(x+1) \over x-1 } | \le 4 |x+1|$.

Now choose $\epsilon>0$ and let $\delta = \min (1, {1 \over 4} \epsilon)$ to get the desired result.