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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.

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    $\begingroup$ I suspect you are not using enough parentheses to express the quantity unambiguously. When in doubt, use extra parentheses! $\endgroup$
    – hardmath
    Sep 8, 2014 at 4:30
  • $\begingroup$ yeah I didn't add them in OP but when I write/input on websites/calculators I always paranthesize the entire f(x)/g(x) $\endgroup$
    – brianforan
    Sep 8, 2014 at 4:31
  • $\begingroup$ Let me edit it for you, and you check my interpretation. $\endgroup$
    – hardmath
    Sep 8, 2014 at 4:32
  • $\begingroup$ $$(2-\sqrt{x+2})(2+\sqrt{x+2})=2^2-(\sqrt{x+2}^2=4-(x+2)=2-x$$ You somehow got $x-6$ instead. Check what went wrong. $\endgroup$ Sep 8, 2014 at 4:33
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    $\begingroup$ A "pro" tip: The numerator $2-\sqrt{x+2}$ evaluates to zero at $x=2$ before the multiplication. Therefore it evaluates to zero after being multiplied by $2+\sqrt{x+2}$. Eventually you want to cancel a factor that makes it zero, but not yet. Use this as a reality check to catch errors like you getting $x-6$ in the numerator. Same with the denominator. $x^2-6x+8$ evaluates to zero also. This is the reason why you started the rationalizing gymnastics in the first place. Therefore it still evaluates to zero after multiplication (but before cancellation). $\endgroup$ Sep 8, 2014 at 4:39

2 Answers 2

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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.

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  • $\begingroup$ yes!! this is where I went wrong, where did the 4- go from 4-(x+2)? I had x-6 because I did 4+2 and was left with 6-x.. which now that I type out doesn't make any sense. I've twisted my potential answer like 5-6 times now over the course of a few hours, in my opinion I'm genuinely trying the question and can't figure it out.. :( $\endgroup$
    – brianforan
    Sep 8, 2014 at 4:38
  • $\begingroup$ $4 - (x + 2) = 4 - x - 2 = 2 - x = -x + 2 = -1(x - 2)$. Does that help? $\endgroup$ Sep 8, 2014 at 4:41
  • $\begingroup$ It's very easy to lose track of signs and make simple mistakes during problems like these. If you have to, write down every single step. Skipping steps can easily lead to mistakes if you haven't practiced much. $\endgroup$ Sep 8, 2014 at 4:43
  • $\begingroup$ I'm just having some basic order of operations problems I think.. which is a little embarrassing, but I'll ask anyway for clarification. You're losing me at $4 - x - 2 = 2- x$, wouldn't it be $4 - -2$ or $4+2$? this is where I'm losing it. I attempted to use formatting in this comment, hoping it works $\endgroup$
    – brianforan
    Sep 8, 2014 at 4:44
  • $\begingroup$ This is the same thing as $4 + (-1)(x + 2)$, which, when we multiply the $-1$ out, is $ 4 + (-x) + (-2) = 4 - x - 2$ $\endgroup$ Sep 8, 2014 at 4:48
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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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  • $\begingroup$ yes but I don't know L'Hospital's rule.. where is the -1/2 coming from and the ^-1/2? It doesn't make any sense to me.. and where did everything except 2x-6 go in the denom? I want to be better with math but it seems like I missed a giant chunk somewhere $\endgroup$
    – brianforan
    Sep 8, 2014 at 4:32
  • $\begingroup$ I reread your question and added an explanation. (I took the derivatives of numerator and denominator.) $\endgroup$
    – John
    Sep 8, 2014 at 4:34
  • $\begingroup$ There are some other restrictions: en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule $\endgroup$
    – John
    Sep 8, 2014 at 4:36
  • $\begingroup$ yeah I get what you're trying to say.. I've read a similar answer on several different websites for different functions. I recognize f(x)/g(x) and f'(x)/g'(x) but when I google "what does a tick mean in f(x)" I don't get any answers because I don't know the formal name for them.. I sort of know what the interminate form is but it doesn't help that I don't know where you got some of those numbers from/got rid of others. $\endgroup$
    – brianforan
    Sep 8, 2014 at 4:41
  • $\begingroup$ I thought that 5-6 hours would be ample time to finish this assignment and that it wouldn't become a night before it's due task, which is my fault, but I guess I'll have to camp out some khan academy videos or something pretty soon $\endgroup$
    – brianforan
    Sep 8, 2014 at 4:42

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