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Square roots are always causing trouble for me - especially finding the limit of a function when it's in the indeterminate form.

I got this far, but I don't know what to do next or even if it's right. Trying to go any further gives me my original problem.

This is the initial problem $$ \lim_{x\to5}\frac{\sqrt{ x^2+5}-\sqrt{30}}{(x-5)} $$

This is where I've gotten $$ \lim_{x\to5}\frac{\sqrt{x^2+5}-\sqrt{30}}{(x-5)} * \frac{\sqrt{x^2+5}+\sqrt{30}}{\sqrt{x^2+5}+\sqrt{30}} = \frac{x^2+5-30}{(x-5)\left(\sqrt{x^2+5}+\sqrt{30}\right)} $$

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I've tidied up the LaTeX in your post, and also fixed a few typos. – Zev Chonoles Aug 22 '11 at 2:30
Can you factor $x^2 - 25$ in any interesting ways? You're almost there. – Dylan Moreland Aug 22 '11 at 2:30
Your final equality should have a $\lim\limits_{x\to 5}$ on the right hand side of the equal sign as well. – Arturo Magidin Aug 22 '11 at 2:34
@Dylan Moreland: I got it down to (x+5)/(sqrt(x^2+5)+sqrt(30)) but the answer is an ugly decimal and according to our teachers we shouldn't be getting those. I don't know if it's something I did wrong or what, but I'm pretty sure it has to either be a whole number or simple fraction. – Joe Aug 22 '11 at 2:56
@Joe: The correct answer to this problem is in fact not a whole number or a simple fraction. It's an expression involving $\sqrt{30}$. You (probably) don't need to give a decimal approximation but can just leave the answer in the form you already have. (More precisely, in your previous comment, you should still be taking the limit as $x \rightarrow 5$. In the form you currently have it this can be done just by substituting $x = 5$, and then you will be done.) – Pete L. Clark Aug 22 '11 at 3:02
up vote 2 down vote accepted

HINT $\ \ \ $ Factoring both of$\ \ \sqrt{x^2+5}^{\:2} - \sqrt{30}^{\:2}\ =\ x^2 - 5^2\ $ by difference of squares yields

$$\frac{\sqrt{x^2+5}-\sqrt{30}}{x-5}\ =\ \frac{x+5}{\sqrt{x^2+5}+\sqrt{30}} $$

Alternatively, if you're familiar with derivatives, note that the limit is $f\:'(5)$ for $f(x) = \sqrt{x^2+5}\:.$ Many limits can be simply calculated by recognizing them as instances of first derivatives and then calculating the derivatively rotely using known derivative rules. You can find a handful of examples in my prior posts starting here. Later in your course you will most likely encounter a generalization of this observation known as l'Hôpital's Rule.

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The problem is (x-5) in the denominator being multiplied by everything else in the denominator. I can't have 0 down there. – Joe Aug 22 '11 at 2:47
Why the downvote? If something is not clear please ask questions and I am happy to explain. – Bill Dubuque Aug 22 '11 at 2:50
@Joe: Have you gotten to derivatives in your calculus class? If not, revisit this comment once you have. Bill's comment turns this standard limit problem into a routine limit problem. – JavaMan Aug 22 '11 at 2:50
I can't vote on answers yet; I wouldn't vote anything with good intent down anyways. @DJC: Not that I recall. School only started for me about a week ago and the first thing we did in calculus were easy limits then we got indeterminates which is where I'm stuck at. – Joe Aug 22 '11 at 2:54

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