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How do we solve this? I am stuck because of sqrt root to the power of 3

$$\lim_{x \to 0} \frac{(1+x)^{1/3}-1}{x}$$

Thsnk you for helping!

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  • $\begingroup$ Hint: $x \mapsto u-1$ and then $u \mapsto v^{\frac13}$ $\endgroup$ – Daniel R Sep 30 '15 at 14:05
  • $\begingroup$ What do you mean by sqrt 3. $\endgroup$ – Display name Sep 30 '15 at 14:43
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Here's a pretty standard technique. Use the identity $(a-b)(a^2+ab+b^2)=a^3-b^3$ to simplify the numerator:

$$\begin{align}\lim_{x \to 0} \frac{(1+x)^{1/3}-1}{x} &= \lim_{x \to 0} \frac{(1+x)^{1/3}-1}{x}\cdot \frac{(1+x)^{2/3}+(1+x)^{1/3}+1}{(1+x)^{2/3}+(1+x)^{1/3}+1} \\ &= \lim_{x \to 0} \frac{(1+x)-1}{x((1+x)^{2/3}+(1+x)^{1/3}+1)} \\ &= \lim_{x \to 0} \frac{x}{x((1+x)^{2/3}+(1+x)^{1/3}+1)} \\ &= \lim_{x \to 0} \frac{1}{(1+x)^{2/3}+(1+x)^{1/3}+1} \\ &= \frac{\lim_{x \to 0} 1}{\lim_{x \to 0} ((1+x)^{2/3}+(1+x)^{1/3}+1)} \\ &= \frac 13\end{align}$$

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  • $\begingroup$ Maybe it will be more convenient to directly substitute $u=(1+x)^{1/3}$? It converts to $\lim_{u\to 1}{\frac{u-1}{u^3-1}}$. $\endgroup$ – Mythomorphic Sep 30 '15 at 14:38
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It is a variation rate for the function $\;u^{\tfrac13}$ aat $u=1$, hence it is $$\lim_{x \to 0} \frac{(1+x)^{1/3}-1}{x}=\frac13u^{-\tfrac23}\Biggr\rvert_{u=1}=\frac13.$$

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As $x \to 0 $

Therefore,

$(1+x)^{\frac{1}{3}}\approx1+x^\frac{1}{3}$, by binomial theorem.

Now its direct.

If you know L'Hôpital rule its also a good or simple way, you may try it.

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  • $\begingroup$ No. The binomial series starts with $1+\frac13 x + \frac1{2!}\frac13(\frac13-1)x^2 + \frac1{3!}\frac13(\frac13-1)(\frac13-2)x^3+…$ $\endgroup$ – Lutz Lehmann Sep 30 '15 at 15:38
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    $\begingroup$ @LutzL If x is very small then, $x^2$ and other greater powers will become almost 0. $\endgroup$ – Display name Sep 30 '15 at 15:49
  • $\begingroup$ Yes, but still, $\sqrt[3]x$ as approximation is wrong. $\endgroup$ – Lutz Lehmann Sep 30 '15 at 18:08

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