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I was given this limit to solve, without using L'Hospital rule. It's killing me !! Can I have the solution please ?

$$\lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}}$$

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Use the series expansion –  pritam Jun 24 '12 at 14:02
    
Using the binomial series expansion, is making it more complicated(or I am doing something wrong) ... –  HackToHell Jun 24 '12 at 14:03

2 Answers 2

up vote 3 down vote accepted

$$ \lim_{x\rightarrow 2}\frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} $$ $$ = \lim_{x\rightarrow 2} \frac{2-\sqrt{2+x}}{2^{\frac{1}{3}}-(4-x)^\frac{1}{3}} \times \frac{2+\sqrt{2+x}}{2+\sqrt{2+x}} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3} }{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}$$ $$ = \lim_{x\rightarrow 2} \frac{2 - x}{-(2-x)} \times \frac{2^{2/3} +2^{1/3}(4-x)^{1/3} +(4-x)^{2/3}}{2 + \sqrt{2+x}}$$ $$= -\frac{3 (2) ^{2/3}}{4} = -\frac{3}{2(2)^{1/3}}$$

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Ah, using the a^3 - b^3 expansion, I get the answer as -3/4 2^(2/3) :) :) –  HackToHell Jun 24 '12 at 14:06
1  
wolframalpha seems to agree with it –  Santosh Linkha Jun 24 '12 at 14:12
    
Weeee, wolframalpha can do calculus, easy way to cross check answers :) :) :) :) –  HackToHell Jun 24 '12 at 14:14

The solution below may come perilously close to the forbidden L'Hospital's Rule, though the Marquis is not mentioned.

To make things look more familiar, change signs, and then divide top and bottom by $x-2$. The expression we want the limit of becomes $$\frac{\sqrt{2+x}-2}{x-2} \cdot \frac{x-2}{(4-x)^{1/3}-2^{1/3}}.$$

We recognize the first part of the product as the "difference quotient" $\frac{f(x+a)-f(a)}{x-a}$ where $f(x)=\sqrt{2+x}$ and $a=2$.

We recognize the second part of the product as the reciprocal of the difference quotient $\frac{g(x+a)-g(a)}{x-a}$ where $g(x)=(4-x)^{1/3}$ and $a=2$.

Now take the derivatives.

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Nice Idea :) :) –  HackToHell Jun 24 '12 at 14:41

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