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The question is:

$$\lim_{x\rightarrow0} \frac {(x+1)^{1/3}-1} x$$

Obviously I have to rationalize it such that the x in the denominator is removed, but I'm not sure how to do so. Thanks.

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Hint: What is the value of derivative of $\sqrt[3]{x+1}$ at $x=0$? – fermesomme Jun 8 '14 at 11:01

another way could be for you to use a substitution $x+1=t^3$. Then as $x \to 0$, $t \to 1$. Use this and simplify.

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How would I simplify the x in the denominator (that's the main problem I'm having). – Ash Jun 8 '14 at 0:25
using the substitution I mentioned you get $\lim_{t \to 1}\frac{t-1}{t^3-1}$. Now use the factorization $t^3-1=(t-1)(t^2+t+1)$. – Anurag A Jun 8 '14 at 0:26

I'm assuming you've studied differentiation. Let $f(x)=(x+1)^{1/3}$. Now consider $f'(0)$. By definition,

\begin{align} f'(0)&= \lim_{h\to0}\frac{f(h)-f(0)}{h} \\ &=\lim_{h\to0} \frac{(h+1)^{1/3}-1}{h} \end{align}

This is precisely the limit you want to evaluate. Now $f'(0)$ can be evaluated using your usual rules of differentiation.

$$ f'(x)=\frac{1}{3}(x+1)^{-2/3}$$ and so $f'(0)=1/3$.

Note: This trick can be used to evaluate limits that looks of the form 'difference quotient'. I think the point of the problem is for you to see that the given limit is the definition for $f'(0)$.

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Let's put $1 + x = y$ so that $y \to 1$ as $x \to 0$. Then we have $$\begin{aligned}L &= \lim_{x \to 0}\frac{(1 + x)^{1/3} - 1}{x}\\ &= \lim_{y \to 1}\frac{y^{1/3} - 1}{y - 1}\\ &= \lim_{z \to 1}\frac{z - 1}{z^{3} - 1}\text{ by putting }y = z^{3}, z\to 1\\ &= \lim_{z \to 1}\frac{1}{z^{2} + z + 1} = \frac{1}{3}\end{aligned}$$

Note that we can prove the more general result in the same manner (i.e. using algebraic manipulation) that $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag {1}$$ if $a > 0$ and $n$ is rational. The result holds even if the number $n$ is irrational, but then the proof is not based on algebraical manipulation but rather on theory of logarithm and exponential functions.

Some answers have used the differentiation technique to get the answer. However to prove that $(x^{n})' = nx^{n - 1}$ we do need to use the limit result (1) I have mentioned above so that the reasoning based on differentiation to calculate this is kind of circular and should be avoided unless the question is kind of multiple choice type (where you have to get the answer and not show the working).

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Use L'Hôpital's rule: $\lim\limits_{x\rightarrow0} \left[ \frac {(x+1)^{1/3}-1}{x} \right]=\lim\limits_{x\rightarrow0}\left[\frac{\frac{d}{dx}[(x+1)^{1/3}-1]}{\frac{d}{dx}[x]} \right]=\lim\limits_{x\rightarrow0}\left[\frac{\frac{1}{3}(x+1)^{-2/3}-0}{1}\right]=\lim\limits_{x\rightarrow0}\left[\frac{1}{3}(x+1)^{-2/3}\right]=...$

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