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Find the following limit,

$$\lim_{x\to 0} \frac {\sqrt [5] {x-\tan^{-1} x}}{x^{3/5}}$$

Observation: Both top and bottom are 0 for $\lim_{x\to 0}\sqrt [5] {x-\tan^{-1} x}$ and $\lim_{x\to 0}{x^{3/5}}$. Therefore, this seems to hint that I should use the L'Hopital Rule...

So I try it. $$\lim_{x\to 0} \frac {\sqrt [5] {x-\tan^{-1} x}}{x^{3/5}}=\lim_{x\to 0}\frac {(\frac{1}{5})({x-\tan^{-1} x})^{-4/5}(1-\frac{1}{x^2+1})}{(\frac{3}{5})x^{-2/5}}=\lim_{x\to 0}\frac {(x)^{2/5}(1-\frac{1}{x^2+1})}{3{({x-\tan^{-1} x})^{4/5}}}$$

I notice if I apply L Hopital rule again, I will get back to square one.

After that I tried several ways that included rationalizing the function, etc.

Any hints? Thanks in advance! List them as solutions. I am looking for hints only.

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Note that $$\frac{\sqrt[5]{x-\tan^{-1}x}}{x^{3/5}}=\frac{\sqrt[5]{x-\tan^{-1}x}}{\sqrt[5]{‌​x^3}}=\sqrt[5]{\frac{x-\tan^{-1}x}{x^3}},$$ and continuity lets the limit pass through the radical. – Cameron Buie Oct 24 '12 at 17:54
up vote 4 down vote accepted

Hints: (i) Forget about fifth roots for a while, study $\dfrac{x-\tan^{-1} x}{x^3}$.

(ii) Then using L'Hospital's Rule is fine, but look carefully after you have applied it once.

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