# Proving limit using Epsilon-Delta definition.

Need to prove that $$\lim_{x \to 1} \frac{x^3 - 1}{(x-1)(x-2)} = -3$$ using the $$\epsilon- \delta$$ definition of the limit.

Here is my proof so far:

Let $$\epsilon > 0$$. Need to show that $$\exists \delta > 0$$, such that if $$|x-1|< \delta$$, $$|\frac{x^3 - 1}{(x-1)(x-2)} + 3| < \epsilon$$.

After manipulating the expression, we see that $$|f(x) + 3| = \frac{|x-1||x+5|}{|x-2|}$$.

I'm having difficulty in picking an appropriate $$\delta > 0$$ such that the above expression is smaller than $$\epsilon$$. I think I have to pick $$\delta > 0$$ so that the $$|x+5|$$ and $$1/|x-2|$$ terms are less than some numerical values respectively. I'm just not sure how to do it.

Any advice would be useful!

Let $$\delta = \min(0.5, \frac{\epsilon}{13})$$,
then $$x \in (0.5, 1.5)$$, hence we can conclude that $$x-2 \in (-1.5, -0.5)$$, that is $$|x-2| \in (0.5, 1.5)$$.
Also, $$|x+5| \in (5.5, 6.5)$$.
Hence $$\frac{|x+5|}{|x-2|} \le \frac{6.5}{0.5}=13$$ .
Now, we should be able to bound $$|f(x)+3|$$.