1
$\begingroup$

How do I solve this limits question:

$$\lim_{x\to-2}\frac{4-x^2}{\sqrt{x^2-x-2}-\sqrt{2-x}}$$

I've already factorised the top and the bottom as far as I can but can't seem to reach the answer.

$$ = \lim_{x\to-2}\frac{(2+x)(2-x)}{\sqrt{(x-2)(x+1)}-\sqrt{2-x}}$$

$$ = \lim_{x\to-2}\frac{(2+x)(2-x)}{\sqrt{-(2-x)(x+1)}-\sqrt{2-x}}$$

$$ = \lim_{x\to-2}\frac{(2+x)(2-x)}{\sqrt{2-x}(\sqrt{-(x+1)}-1)}$$

$$ = \lim_{x\to-2}\frac{(2+x)\sqrt{2-x}}{(\sqrt{-(x+1)}-1)}$$

Am I moving in the right direction?

The answer to the question is 4.

$\endgroup$
2
$\begingroup$

Multiply the numerator and denominator by the conjugate of the denominator: multiply your function by $$\frac{\sqrt{x^2-x-2}+\sqrt{2-x}}{\sqrt{x^2-x-2}+\sqrt{2-x}}$$

Recall how a difference of squares factors: $$(a - b)(a+b) = a^2 - b^2$$

In this case, we see that when we multiply numerator and denominator by the conjugate of the denominator, w have a difference of squares in the denominator, which then simplifies greatly:

$$\begin{align} (\sqrt{x^2-x-2}-\sqrt{2-x})(\sqrt{x^2-x-2}+\sqrt{2-x}) & = (x^2 - x - 2) - (2 - x) \\ \\ &= x^2-4 \\ \\ &= -(4-x^2)\end{align}$$

After canceling the common factor of $4 - x^2$ we are left with $$\lim_{x\to -2} -(\sqrt{x^2-x-2}+\sqrt{2-x}) = -4$$

$\endgroup$
  • 1
    $\begingroup$ I think you meant to write $-(4-x^2)$ on the last line. $\endgroup$ – user71641 Jan 20 '14 at 18:01
0
$\begingroup$

Where you have left of $$I=\lim_{x\to-2}\frac{(2+x)\sqrt{2-x}}{(\sqrt{-(x+1)}-1)}$$

Setting $x+2=h$

$$I=\lim_{h\to0}\frac{h\sqrt{4-h}}{(\sqrt{1-h}-1)}$$

$$I=\lim_{h\to0}\frac{h(\sqrt{1-h}+1)}{1-h-1}\cdot \lim_{h\to0}\sqrt{4-h}$$

As $h\to0\implies h\ne0$

$$I=-\lim_{h\to0} (\sqrt{1-h}+1)\cdot\sqrt{2+2}=\cdots $$

$\endgroup$
0
$\begingroup$

Hint $\ -\dfrac{\ \ f\,-\,g}{\sqrt{f}-\sqrt{g}}\, =\, -(\sqrt f + \sqrt{g}).\ $ In your case the RHS has determinate limit.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.