So, I need you to solve one of these limits for me, so I can see how it's done, so I can do the rest myself.
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Or please explain me everything I need to know to solve them :D – A6SE Sep 30 '13 at 18:43
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rationalize the radicals – lab bhattacharjee Sep 30 '13 at 18:44
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@A6Tech This is a homework question. – Sep 30 '13 at 18:52
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It's not, we're doing limits in two classes, I just want to be prepared. The answers below gave me a clue. – A6SE Sep 30 '13 at 19:27
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OK, one. Here comes: $$ \frac{\sqrt{u}-\sqrt{v}}{\sqrt[3]{u}-\sqrt[3]{v}} = \frac{u-v}{\sqrt{u}+\sqrt{v}}\cdot\frac{\sqrt[3]{u^2}+\sqrt[3]{uv}+\sqrt[3]{v^2}}{u-v}= \frac{\sqrt[3]{u^2}+\sqrt[3]{uv}+\sqrt[3]{v^2}}{\sqrt{u}+\sqrt{v}}\underset{u,v\to1}{\longrightarrow}\frac{1+1+1}{1+1}. $$ Equivalently, when $x\to0$, $(1\pm x)^a=1\pm ax+o(x)$ hence $$ \frac{(1+x)^a-(1-x)^a}{(1+x)^b-(1-x)^b}=\frac{1+ax-(1-ax)+o(x)}{1+bx-(1-bx)+o(x)}=\frac{2ax+o(x)}{2bx+o(x)}\underset{x\to0}{\longrightarrow}\frac{a}b. $$
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You could multiply by the conjugate over the conjugate (or the appropriate generalization for the nth roots), and do some asymptotics.
eg: $\lim\limits_{x\rightarrow\infty} (x- \sqrt{x^2-5x} * \frac{x+\sqrt{x^2-5x}}{x+\sqrt{x^2-5x}})$
$= \lim\limits_{x\rightarrow\infty} \frac{x^2 - x^2+5x}{x+\sqrt{x^2-5x}}$ $= \lim\limits_{x\rightarrow\infty} \frac{5x}{x+\sqrt{x^2-5x}}$
At this point, notice that $\lim\limits_{x\rightarrow\infty} x+\sqrt{x^2-5x} =\lim\limits_{x\rightarrow\infty} 2x $ and substitute, to get $5/2$ as the answer.
Chris Bonnell
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Just for the sake of completeness, to figure out "the appropriate generalization" that i mention, Consider that x^2-y^2 = (x-y)(x+y), and the generalization you want is x^n-y^n = (x-y)(cyclotomic polynomial of degree n-1 in x and y). It can be derived from http://en.wikipedia.org/wiki/Cyclotomic_polynomial and then you replace the values with the roots you want. It does get ugly, but it works. – Chris Bonnell Sep 30 '13 at 18:58