I've encountered an interesting question that involves proving the limit of a function of the aforementioned nature. Namely, I have been asked to prove
$\lim \limits_{x \to 1^+}$ $\sqrt(x^3-1)=0$
With my basic knowledge of how to prove similar questions, I have attempted to prove the above by expressing the RHS in terms of $\delta$. However, I have run into a snag whilst doing so. I would like MSE's advice on how to circumvent this issue.
Here's my attempt:
For any $\epsilon >0$, we need to find a $\delta >0$ so that:
$ 1< x <1+\delta \implies \vert\sqrt(x^3-1) \vert<\epsilon$
Considering RHS:
$\vert\sqrt(x^3-1) \vert $ =$ |x^3-1| \over \vert\sqrt(x^3-1) \vert$ = $ |x-1||x^2+x+1| \over \vert\sqrt(x^3-1) \vert$ < $|x-1||(x-1)(x-2)+3|$
(The last inequality has been asserted on the basis of the fact that $\vert\sqrt(x^3-1) \vert $>1)
$|x-1||(x-1)(x-2)+3|$ < $|x-1| (|x-1|+|(x-2)+3|)$
Here, I have applied the triangle inequality. My aim is to express the inequality in terms of $\delta$.
However, I now realize that this application is incorrect. Thus, how else could I go about with this proof? Is there some other way we can squeeze $\delta$ out?
Thank you.
See, it is not a sin to work backwards, as long as your answer is correct in the end.
Note that: $$ \sqrt{x^3-1} < \epsilon \iff x^3 < \epsilon^2 + 1 \iff x < \sqrt[3]{\epsilon^2+1} $$
Now, scratch out what I wrote earlier, and start:
Let $\delta = \sqrt[3]{\epsilon^2+1} - 1$, then $\delta > 0$.Furthermore, $$ x < 1 + \delta \implies x < \sqrt[3]{\epsilon^2+1} \implies \sqrt{x^3-1} < \epsilon $$
Thus, we found our $\delta$, proving the proposition. Don't be afraid to work backwards, it works (and don't reveal it, make the $\delta$ come by magic).