1
$\begingroup$

I need to find the limit of the expression: $$ \lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2} $$

It is given that $b>2$. In the class we were encouraged to use the method of multiplying both the numerator and the denominator with the numerator when the limit can't be found simply by using limit arithmetic. Hence: $$ \lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2} =\\ =\lim_{x\to b}\frac{(\sqrt{b-2} - \sqrt{x-2})(\sqrt{b-2} - \sqrt{x-2})}{(x^2-b^2)(\sqrt{b-2} - \sqrt{x-2})} = \\ = \lim_{x\to b}\frac{b+x-4 - 2\sqrt{b+x-4}}{(x-b)(x+b)(\sqrt{b-2} - \sqrt{x-2})} $$

I don't see though how from this point this can be simplified. We could factor out the $\sqrt{b+x-4}$ in the numerator but this doesn't really help. Wolfram says the answer is $\frac{1}{4b\sqrt{b-2}}$

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Multiply by $\frac{\sqrt{b-2} + \sqrt{x-2}}{\sqrt{b-2} + \sqrt{x-2}}$ $$\lim_{x\to b}\frac{\sqrt{b-2} - \sqrt{x-2}}{x^2-b^2}\frac{\sqrt{b-2} + \sqrt{x-2}}{\sqrt{b-2} + \sqrt{x-2}}$$ $$\lim_{x\to b}\frac{b-x}{(x-b)(x+b)(\sqrt{b-2} + \sqrt{x-2})}=\frac{-1}{(b+b)(\sqrt{b-2}+\sqrt{b-2})}=\frac{-1}{4b\sqrt{b-2}}$$

$\endgroup$
4
  • $\begingroup$ Thank you! But in this transition $lim_{x\to b} \frac{-1}{(x+b)(\sqrt{b-2}+\sqrt{x-2)}}$ how do we get to $4b\sqrt{b-2}$ in the denominator? $\endgroup$
    – Yos
    Nov 19, 2016 at 11:30
  • $\begingroup$ @Yos I have added more details $\endgroup$
    – E.H.E
    Nov 19, 2016 at 11:36
  • $\begingroup$ When we have $\frac{b-x}{(x-b)(x+b)}$, doesn't it become $\frac{-1}{x+b}$? $\endgroup$
    – Yos
    Nov 19, 2016 at 11:40
  • $\begingroup$ @Yos yes and then let $x\rightarrow b$ $\endgroup$
    – E.H.E
    Nov 19, 2016 at 11:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .