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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!

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1 Answer 1

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One common trick is to use the minimum function.

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 $|f(x)+3|$.

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