How do I solve this limits question:


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.


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 $$\lim_{x\to -2} -(\sqrt{x^2-x-2}+\sqrt{2-x}) = -4$$

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

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

Setting $x+2=h$


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

As $h\to0\implies h\ne0$

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


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


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