I'm extremely new to calculus so please excuse my lack of lingo/formatting.
I'm doing homework for my calc class, and I looked on wolfram alpha. It told me the limit was $\frac{1}{8}$ but I wanted to do it on my own to make sure I actually knew what I was doing. Wolfram$\alpha$ told me to use l'Hospital's rule, but I've never learned it and couldn't figure it out based on some google searches. It also said the limit as $x\rightarrow2$ was $\frac 1 8$, but the answers I got are either $-\frac {1} {8}$ or $-1$.
Here is the problem:
$$ \lim_{x\to 2} \frac{2 - \sqrt{x+2}}{ x^2 - 6x + 8} $$
So I tried to rationalize by multiplying the numerator by $2 + \sqrt{x+2}$, but then my final answer came out to $\frac{-4}4$ when I plugged $2$ into
$$ \frac{x-6}{(x^2-6x+8)(2+\sqrt{x+2})}$$
I'm really just not sure what I'm doing wrong. I haven't taken a precalc course since senior year and I'm a sophomore now, but we did mostly trig, so derivatives and all that are absolutely new to me. Any help is appreciated.
$\endgroup$ 62 Answers
$\begingroup$I'm assuming you mean $$\lim_{x \to 2} \frac{2 - \sqrt{x + 2}}{x^2 - 6x + 8}$$
Just multiply by the conjugate: $$\lim_{x \to 2} \frac{2 - \sqrt{x + 2}} {x^2 - 6x + 8} \times 1 = \lim_{x \to 2} \frac{2 - \sqrt{x + 2}} {x^2 - 6x + 8} \times \frac{2 + \sqrt{x + 2}}{2 + \sqrt{x + 2}} = \lim_{x \to 2} \frac{4 - (x + 2)}{(x^2 - 6x + 8)(2 + \sqrt{x + 2})} = \lim_{x \to 2}\frac{-(x - 2)}{(x - 4)(x - 2)(2 + \sqrt{x + 2})}$$
Now we can cancel out the $(x - 2)$ to get $$\lim_{x \to 2} \frac{-1}{(x - 4) (2 + \sqrt{x + 2})} = \frac{-1}{(-2)(2 + \sqrt{4})} = \frac{-1}{(-2)(4)} = \frac{1}{8}$$
Your best strategy when you see a square root in a context like this is to multiply by the conjugate and just play around with the expression, simplifying what you can until it's no longer in an indeterminate form.
$\endgroup$ 6 $\begingroup$You can use L'Hospital's rule:
$$\lim_{x \to 2} \frac{2 - (\sqrt{x+2)}}{x^2 - 6x + 8} =$$ $$\lim_{x \to 2} \frac{-\frac{1}{2}(x+2)^{-\frac{1}{2}}}{2x - 6} = \frac{-\frac{1}{4}}{-2} = \frac{1}{8}.$$
The form of the limit is indeterminate $\frac{0}{0}$ so
$$\lim_{x \to 2} \frac{f(x)}{g(x)} = \lim_{x \to 2} \frac{f'(x)}{g'(x)},$$
assuming that the right side isn't itself an indeterminate form (which it isn't).
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