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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!

zeta1203
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3 Answers3

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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 $|f(x)+3|$.

Siong Thye Goh
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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 $$0<|x-1|<\delta\implies |f(x) +3|<\epsilon$$

Paramanand Singh
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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}

So_M.
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