# 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.

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 $$|f(x)+3|$$.

Your thinking and approach is correct. Also you can observe that bounding $$|x+5|$$ is not the main issue because if $$x$$ is near $$1$$ then $$|x+5|$$ should be near $$6$$ (just put values of $$x=0.9,1$$ etc).

The more difficult thing to handle is $$1/|x-2|$$, but notice that this will be become large if $$x-2$$ becomes small. In other words we need to keep $$x$$ near $$1$$ and a bit far from $$2$$. Mid point of $$1$$ and $$2$$ is $$1.5$$ and its distance from each is $$1/2$$.

Thus if $$|x-1|<1/2$$ we have $$|x-2|>1/2$$ ie $$1/|x-2|<2$$. Also under same condition we have $$|x+5|\leq |x-1|+6<7$$ and thus we have the implication $$|x-1|<1/2\implies \frac{|x+5|}{|x-2|}<14$$ so that $$|f(x) +3|<14|x-1|$$ and thus if we set $$\delta=\min(1/2,\epsilon/14)$$ we have the desired implication $$0<|x-1|<\delta\implies |f(x) +3|<\epsilon$$

Let $$\epsilon>0$$, and let $$\delta:=\dfrac{\epsilon}{7+\epsilon}.$$ Note that simple calculation shows $$0<\delta<1,$$ hence $$\delta^2\leqq\delta$$, and by triangle inequality, we get $$|x-2|=|1-(x-1)|\geqq 1-|x-1|.$$

Then, supposing $$0<|x-1|<\delta$$, we have \begin{align} \left|\frac{x^3-1}{(x-1)(x-2)}+3\right| &=\left|\frac{(x-1)(x+5)}{x-2}\right|\\ &=\frac{|(x-1)^2+6(x-1)|}{|x-2|}\\ &\leqq\frac{|(x-1)^2+6(x-1)|}{1-|x-1|}\\ &\leqq\frac{|x-1|^2+6|x-1|}{1-|x-1|}\\ &\leqq\frac{\delta^2+6\delta}{1-\delta}\\ &\leqq\frac{\delta+6\delta}{1-\delta}\\ &=\frac{7\delta}{1-\delta}\\ &\leqq\frac{7\cdot\tfrac{\epsilon}{7+\epsilon}}{1-\tfrac{\epsilon}{7+\epsilon}}\\ &=\epsilon. \end{align}