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I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$

Could you give me a hint how to start solving this? (I know the result is $3$)

Thanks a lot !

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As you've written it, the limit is 0. The function is defined at x=1, so you don't even need to bother with taking the limit. –  Michael Dyrud Sep 24 '12 at 18:33
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Is it suppose to be cubic root in your denominator? –  Jean-Sébastien Sep 24 '12 at 18:33
    
I'm afraid you made a mistake as the limit exists and it's zero, not three. Check this. –  DonAntonio Sep 24 '12 at 18:36
    
@DonAntonio the mistake was the missing 3 on the root ;) –  foaly Sep 24 '12 at 18:38
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@foaly: All of us are here to share our knowledge about Maths. Some Masters like André and others help you and me. It will be a good sign to others that you take the best answer or vote them up. Try to do what you did mints ago for all you previous questions and then you will see your accept rate will increase. :-) –  B. S. Sep 26 '12 at 14:00
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5 Answers

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You know that $$(x^3-y^3)=(x-y)(x^2+xy+y^2)$$ and $$(a^2-b^2)=(a-b)(a+b)$$ so if $a=\sqrt{3+x}$ and $b=2$, we have $(a^2-b^2)=(\sqrt{3+x}-2)(\sqrt{3+x}+2)=3+x-4$. The same calculation can be done for another identity by taking $x=\sqrt[3]{7+x}$ and $y=2$. In fact you have $$(\sqrt[3]{7+x}-2)\big((\sqrt[3]{7+x})^2+2\sqrt[3]{7+x}+2^2\big)=(\sqrt[3]{7+x})^3-2^3=7+x-8$$ Now multiply your fraction to $\frac{\sqrt{3+x}+2}{\sqrt{3+x}+2}=1$ and $\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}=1$ simultaneously, so: $$\frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}=\frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}\times\frac{\sqrt{3+x}+2}{\sqrt{3+x}+2}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}$$ $$=\frac{3+x-4}{7+x-8}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2}$$ $$=\frac{x-1}{x-1}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2} $$ which is equal to $$\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2}$$ near $x=1$. Now, I think taking the limit when $x$ tends to $1$ is so easy. It is $3$.

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i'm afraid i can't follow you.. pretty much from the beginning. –  foaly Sep 27 '12 at 8:19
    
@foaly: Where are you stuck? –  B. S. Sep 27 '12 at 9:06
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thank you very much. shouldn't it be $$(a^2-b^2)=(\sqrt{3+x}-2)(\sqrt{3+x}+2)=3+x-4$$ though? Also: i can comprehend your doing now, but not necessarily reproduce. How do you start solving something like this, and what would i do if it was a much higher grade root? –  foaly Sep 27 '12 at 10:54
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@foaly: I hope I could help. I think for higher grade root maybe André 's and Israel's would be useful but for small grade my way seems practical. :-) –  B. S. Sep 27 '12 at 20:37
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Use L'Hospital's Rule.

If we want to avoid explicit mention of L'Hospital's Rule, rewrite our expression as $$\frac{\sqrt{3+x}-2}{x-1}\frac{x-1}{\sqrt[3]{7+x}-2},$$ and observe that $$\lim_{x\to 1}\frac{\sqrt{3+x}-2}{x-1}\quad\text{and}\quad \lim_{x\to 1}\frac{\sqrt[3]{7+x}-2}{x-1}$$ are each derivatives.

If we want completely to avoid using a rule of differentiation, we can calculate the limits by the usual algebra tricks.

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oh ya sorry ~ i'll edit. –  foaly Sep 24 '12 at 18:36
    
I didn't hear about the l'hopital's rule yet, and in the book its way behind this exercise. is it necessary to use that rule? –  foaly Sep 24 '12 at 18:41
    
I gave a different argument below the L'Hospital's Rule one. –  André Nicolas Sep 24 '12 at 18:51
    
i'll look at it tmr.. gotta sleep now x.x –  foaly Sep 24 '12 at 18:55
    
okay... i don't know why they are 'each derivatives', nor do i know why you can rewrite it as $$\frac{\sqrt{3+x}-2}{x-1}\frac{x-1}{\sqrt[3]{7+x}-2}$$.. –  foaly Sep 27 '12 at 8:22
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For convenience, change variables to $t = x - 1$, so you're looking at $\dfrac{\sqrt{4+t}-2}{(8+t)^{1/3}-2}$.

$$ \sqrt{4+t} = 2 \sqrt{1+\frac{t}{2}} = 2 \left(1 + \frac{t}{4} + \ldots\right) $$ $$ (8+t)^{1/3} = 2 \left(1 + \frac{t}{8}\right)^{1/3} = 2 \left(1 + \frac{t}{24} + \ldots\right) $$

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i'll look at it tmr.. gotta sleep now x.x –  foaly Sep 24 '12 at 18:53
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The method totally depends on how much calculus you have studied. l'hospitals rule makes life easier in most cases but in this case in this case l'hospitals rule will be tedious since to take derivatives of square roots and cuberoots is a tad bit tedious as compared to rationalisation. so i would prefer rationalisation such that we get a rational number in the denominator. also l'hospitals rule is quite sophisticated for such simple problems, and to use it is similiar to differetiating a quadratic equation to find its minima and maxima instead of just completing the squares.

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$$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}=\lim_{x\to 1} \frac{\sqrt{3+x}-2}{x-1}\cdot \lim_{x\to 1} \frac{x-1}{\sqrt[3]{7+x}-2}$$ $$=\lim_{x+3\to 4} \frac{{(x+3)^{0.5}}-4^{0.5}}{(x+3)-4}\cdot \frac{1}{\lim_{(x+7)\to 8}\frac{({x+7})^{1/3}-8^{1/3}}{(x+7)-8}}$$

Use $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$$ which gives

$$\lim_{x+3\to 4} \frac{{(x+3)^{0.5}}-4^{0.5}}{(x+3)-4}\cdot \frac{1}{\lim_{(x+7)\to 8}\frac{({x+7})^{1/3}-8^{1/3}}{(x+7)-8}}=(0.5)4^{-0.5}\cdot\frac{1}{(1/3)8^{-2/3}}=3$$

see

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