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I have problem with my limit set from my lecture, I don't know how to get off from this problem. Please help me.

$$\lim_{t\to0}\frac{\sqrt[3]{t+1} - 1}{t}$$

Please help me, i need your help, thanks.

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By "sqrt3", I assume you meant a cube root? Your link appears not to be working. – Matthew Pressland Oct 9 '12 at 15:17
the link was edited, please check the link. thanks – NPE Oct 9 '12 at 15:22
up vote 1 down vote accepted

Define $f: \mathbb{R}\to\mathbb{R}$ by $f(x)=x^{1/3}$, $f'(x)=\frac{1}{3 x^{2/3}}$then \begin{equation} \lim_{t\to 0}\frac{(t+1)^{1/3}-1}{t}=\lim_{t\to 0}\frac{f(t+1)-f(1)}{t}=f'(1)=\frac{1}{3}. \end{equation}

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thank you very much,, – NPE Oct 9 '12 at 15:40

I can't resist throwing the algebraic version into the pot: Start with the identity $$(a-b)(a^2+ab+b^2)=a^3-b^3,$$ put $a=\sqrt[3]{t+1}$ and $b=1$, and get $$(\sqrt[3]{t+1}-1)(a^2+ab+b^2)=t.$$ Divide by $t$ and let $t\to0$, and note that then $a^2+ab+b^2\to3$.

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Using the L'Hôpital's rule: $$ \lim_{t\to0}\frac{({t+1})^{1/3} - 1}{t}\overset{l'Hopital}{=}\lim_{t\to0}\frac{1/3({t+1})^{-2/3}}{1}=\frac13 $$

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thank you very much, I've got what I was looking for. – NPE Oct 9 '12 at 15:40

$$\lim_{t\to 0}\frac{\sqrt[3]{t+1} - 1}{t}=$$

$$=\lim_{t\to0}\frac{\sqrt[3]{t+1} - 1}{t}\frac{\sqrt[3]{(t+1)^2}+\sqrt[3]{t+1}+1^2}{\sqrt[3]{(t+1)^2}+\sqrt[3]{t+1}+1^2}=$$ $$=\lim_{t\to0}\frac{\sqrt[3]{(t+1)^3} - 1^3}{t(\sqrt[3]{(t+1)^2}+\sqrt[3]{t+1}+1^2)}=$$ $$=\lim_{t\to0}\frac{(t+1) - 1}{t(\sqrt[3]{(t+1)^2}+\sqrt[3]{t+1}+1^2)}=$$ $$=\lim_{t\to0}\frac{t}{t(\sqrt[3]{(t+1)^2}+\sqrt[3]{t+1}+1^2)}=$$ $$=\lim_{t\to0}\frac{1}{\sqrt[3]{(t+1)^2}+\sqrt[3]{t+1}+1}=\frac{1}{3}$$

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Your teacher may be looking to have you do this without derivatives, in which case what you want is the binomial expansion: $\displaystyle \left(1+t\right)^x = 1+xt+\frac{x(x-1)t^2}{2!}+\frac{x(x-1)(x-2)t^3}{3!}+\cdots$ (Note that this formula works even if $x$ isn't an integer - it's just an infinite sum rather than a finite one, in that case, and you have to worry about convergence of the series, but those issues don't arise here.) You should be able to plug this into your formula, and doing the algebra you'll see that the higher-order terms all still contain factors of $t$ and thus go to zero as $t\rightarrow 0$.

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is the answer 1/3? – NPE Oct 9 '12 at 15:42
@john The result is by far the least important part of the problem, which is why I left it to you to figure out (though certainly four other answers have told you the number!). The answer isn't 1/3; the answer is the process that you use to get the number. – Steven Stadnicki Oct 9 '12 at 15:59

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