# Thread: Limits - Solve By Manipulation

1. ## Limits - Solve By Manipulation

Okay, so I know you have to rationalize the limit in order to get it out of the indeterminate form, but I can't seem to do it for these two questions. Can someone help, start me off?

a) Lim as x -> 4 of (x^2-4)/(sqrtx - 2)

b) Lim as x -> 0 of [sqrt(3-x) - sqrt(x+3)] / x

Thanks,
Stefan

2. a) $\lim_{x \to 4} \frac{x^2 - 4}{\sqrt{2}-2} = \left[\frac{12}{0}\right]^+$

This is not an indeterminate form. You'll find that the limit does not exist. Consider the left and right handed limits and you'll see they're not equal.

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b) \begin{aligned} \lim_{x \to 0} \frac{\sqrt{3-x} - \sqrt{x+3}}{x} \cdot {\color{blue} \frac{\sqrt{3-x} + \sqrt{x+3}}{\sqrt{3-x} + \sqrt{x+3}}}\\ = \lim_{x \to 0} \frac{3-x - (x+3)}{x\left(\sqrt{3-x}+\sqrt{x+3}\right)} \\ = \lim_{x \to 0} \frac{-2x}{x\left(\sqrt{3-x}+\sqrt{x+3}\right)} \\ \end{aligned}

Notice that something cancels Then you can plug in $x = 0$ without getting an indeterminate form.

3. Hi, thanks or the help, a quick question though. When we rationalize why do we use the opposite sign, for example, (b) had sqrt + sqrt and when it was rationalized it was sqrt - sqrt. What is the actual reason this is done, I know it cancels but what proof is there that that always works?

4. Well, it all boils down to the difference of squares formula: $(a-b)(a+b) = a^2 - b^2$

In particular, if $a$ and $b$ are square roots, we would like to get rid of them and by multiplying by the conjugate (in this case, $(a+b)$) we get rid of them and we get an expression without square roots, i.e. $a^2 - b^2$