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$$\lim_{x \to \pi/2} \frac{\sqrt[4]{ \sin x} - \sqrt[3]{ \sin x}}{\cos^2x}$$

I have an idea of replacing $\sin x$ to $n$ when $n \to 1$ but wolfram says that answer is $\frac{\pi}{48} $ so my suggestion is it's had to use trigonometry simplifications which I do not know so well.

Assuming that L'Hopital is forbidden but you can use asimptotical simplifications like big and small o notations and Taylor series.

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Let $\sin x=n^{12}$ then your problem becomes:

$$\lim_{n\rightarrow1}\frac{n^3-n^4}{1-n^{24}}$$

$$=\lim_{n\rightarrow1}\frac{n^3(1-n)}{(1-n)\sum_{i=0}^{23}n^i}$$

$$=\lim_{n\rightarrow1}\frac{n^3}{\sum_{i=0}^{23}n^i}$$

$$=\frac{1}{24}$$

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  • $\begingroup$ Yeah, I've done that already. Awesome replacement was an awesome hint. $\endgroup$ – Val Nov 18 '15 at 8:14
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Ian Miller's answer is the nicest and most efficient solution to the problem.

Just for your curiosity, I shall give you another one using Taylor series since you will use them a lot during your studies.

First, changing variable $x=\frac \pi 2+y$ $$\frac{\sqrt[4]{ \sin (x)} - \sqrt[3]{ \sin (x)}}{\cos^2(x)}=\frac{\sqrt[4]{ \cos(y)} - \sqrt[3]{ \cos (y)}}{\sin^2(y)}$$ Now, using the generalized binomial expansion and Taylor series around $y=0$ $$cos^a(y)=1-\frac{a y^2}{2}+\left(\frac{a^2}{8}-\frac{a}{12}\right) y^4+O\left(y^6\right)$$ Using it, the numerator write $$\frac{y^2}{24}+\frac{y^4}{1152}+O\left(y^6\right)$$ and the denominator is $\approx y^2$. Then, the limit.

We could go further and use $$\sin(y)=y-\frac{y^3}{6}+O\left(y^4\right)$$ So the denominator is $$y^2-\frac{y^4}{3}+O\left(y^5\right)$$ which makes $$\frac{\sqrt[4]{ \cos(y)} - \sqrt[3]{ \cos (y)}}{\sin^2(y)}=\frac{\frac{y^2}{24}+\frac{y^4}{1152}+O\left(y^6\right) }{y^2-\frac{y^4}{3}+O\left(y^5\right) }$$ and the long division yields to $$\frac{\sqrt[4]{ \cos(y)} - \sqrt[3]{ \cos (y)}}{\sin^2(y)}=\frac{1}{24}+\frac{17 }{1152}y^2+O\left(y^3\right)$$ which shows the limit and also how it is approached.

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