The following limit exists. Find the value of the limit.
$\displaystyle\lim_{x\to 16} \ \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7 \ } - 3)}{(\sqrt{x - 12 \ } - 2)(\sqrt{x - 15 \ } - 1)}$
(Edit)
The following limit exists. Find the value of the limit. Please show your work.
$\displaystyle\lim_{x\to 16} \ \dfrac{(\sqrt{x} - 4)(\sqrt{x - 7 \ } - 3)}{(\sqrt{x - 12 \ } - 2)(\sqrt{x - 15 \ } - 1)}$
Nice algebra problem. Thanks.
Prove It was on the right track for a formal demonstration of Archie's result. And of course Plato's answer is elegant. For anyone who wants detail, see below:
Spoiler:
Tastes differ. In my opinion, his statement is an elegant consequence of utilizing conjugates and indicates why it is a neat problem in the first place. I do agree that a beginning student is highly unlikely to follow so someone should provide a detailed development (as I did). But that does not detract from its simplicity and clarity (for those familiar with conjugates).
In terms of limits solved using conjugate pairs, I prefer limits of the form
$\displaystyle \lim_{x \to \infty} \left( \sqrt{x^2 + ax + b} - \sqrt{x^2 + cx + d} \right)$because the convergence is, to me, somewhat counter-intuitive.
If memory serves, I recently saw an example that combined the two ideas.
For those of us who have had a life-time of experience teaching/writing about this topic surely realize that students working on these types of limits should have a solid grounding in basic algebra manipulation of radicals. Any of us should have taught the time honored 'trick' of inverse/sign-change. That is all that I posted. It skipped no steps, it left nothing out that someone competent is basic algebra might fail to see.
Here is that trick again.
$\displaystyle \left( \sqrt{x^2 + ax + b} - \sqrt{x^2 + cx + d} \right)=\dfrac{(a-c)x+(b-d)}{\sqrt{x^2 + ax + b} + \sqrt{x^2 + cx + d}}\to\dfrac{a-c}{\sqrt{2}}$ as $x\to\infty$.