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Applying the following expansions:

$\sqrt[]{\cos x}=1-\frac{x^2}{4}+O(x^4)$

$\sqrt[3]{\cos x}=1-\frac{x^2}{6}+O(x^4)$

$\sin^2 x=x^2+O(x^4)$

gives the correct result on the limit $L=\frac{-1}{12}$.

But when applying more orders of each function:

$\sqrt[]{\cos x}=1-\frac{x^2}{4}-\frac{x^4}{96}+O(x^5)$

$\sqrt[3]{\cos x}=1-\frac{x^2}{6}-\frac{x^4}{72}+O(x^5)$

$\sin^2 x=x^2-\frac{x^4}{3}+O(x^6)$

gives $L=\frac{1}{96}$.

So, in general, how to choose the number of terms in Taylor (Maclaurin) series to evaluate the limit?

EDIT:

Applying the higher order than $O(x^4)$:

$$\sqrt[]{\cos x}-\sqrt[3]{\cos x}=1-\frac{x^2}{4}-\frac{x^4}{96}-1+\frac{x^2}{6}+\frac{x^4}{72}=\frac{-x^2}{12}+\frac{x^4}{288}$$

$$\sin^2 x=x^2-\frac{x^4}{3}+O(x^6)=x^2-\frac{x^4}{3}$$

$$\frac{\sqrt[]{\cos x}-\sqrt[3]{\cos x}}{\sin^2 x}=\frac{x^4\left(\frac{-24}{x^2}+1\right)}{96x^4\left(\frac{3}{x^2}-1\right)}\Rightarrow L=\frac{-1}{96}$$

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  • $\begingroup$ How does it give $\frac{1}{96}$? $\endgroup$ – GoodDeeds Feb 3 '16 at 14:12
  • $\begingroup$ @GoodDeeds I will edit the question to show the possible errors. $\endgroup$ – user300048 Feb 3 '16 at 14:18
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    $\begingroup$ @user300044 You'll get $\frac{-1}{12}$. All other terms will vanish as $x \to 0$ $\endgroup$ – Max Payne Feb 3 '16 at 14:24
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    $\begingroup$ $$-\frac{1}{12}$$ is the right result $\endgroup$ – Dr. Sonnhard Graubner Feb 3 '16 at 14:32
  • $\begingroup$ You get $\sum_{n=1}^{\infty} {n} = -\frac{1}{12}$ $\endgroup$ – George Feb 3 '16 at 15:41
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So, in general, how to choose the number of terms in Taylor (Maclaurin) series to evaluate the limit?

To answer your general question about how to approach problems like these, it's useful to know the concept of equivalence of functions. Given two functions $f$ and $g$ defined and nonzero in a neighbourhood of $a$, we write $f \sim g$, and say $f$ and $g$ are equivalent, if $f(x)/g(x) \to 1$ as $x \to a$. This is an equivalence relation. The condition $f \sim g$ is equivalent to $f = g + o(g)$. Importantly, if $f_1 \sim f_2$ and $g_1 \sim g_2$, then $f_1 g_1 \sim f_2 g_2$, and $f_1^{\alpha} \sim f_2^{\alpha}$. Also, $o(f_1) = o(f_2)$ and $O(f_1) = O(f_2)$.

Consequently, any time you are evaluating the limit of a product or quotient of several functions (or powers of functions), your goal should be to replace each factor with a simple equivalent function. For example, if $f \sim f_1, g \sim g_1, h \sim h_1$, then $f^3 g^2/h^{5/3} \sim f_1^3 g_1^2/h_1^{5/3}$.

In your case, you would like to find simple equivalents for the factors $\sin x$ and $(\cos x)^{1/2} - (\cos x)^{1/3}$. In general, a simple equivalent for a function near a point $a$ will be given by the first nonzero term in its Taylor series. Therefore, in the present case where $x \to 0$, you will use $\sin x \sim x$, and you will want to have enough terms of the series for $(\cos x)^{1/2}$ and $(\cos x)^{1/3}$ that you know the first nonzero term in the series for $(\cos x)^{1/2} - (\cos x)^{1/3}$. (Strictly speaking, since your denominator is equivalent to $x^2$ and you only want the limit rather than a simple equivalent, you only need these series up to the second order. But it's best to understand the general principle I've stated.)

We have $$(\cos x)^{1/2} = (1 - x^2/2 + o(x^2))^{1/2} = 1 + \frac{1}{2}(-x^2/2 + o(x^2)) + o(- x^2/2 + o(x^2)) = 1 - x^2/4 + o(x^2),$$ $$(\cos x)^{1/3} = (1 - x^2/2 + o(x^2))^{1/2} = 1 + \frac{1}{3}(-x^2/2 + o(x^2)) + o(- x^2/2 + o(x^2)) = 1 - x^2/6 + o(x^2),$$ hence $$(\cos x)^{1/2} - (\cos x)^{1/3} = -x^2/12 + o(x^2) \sim -x^2/12.$$ Therefore $$\frac{(\cos x)^{1/2} - (\cos x)^{1/3}}{\sin^2 x} \sim \frac{-x^2/12}{x^2} = -1/12.$$ This proves that your limit is $-1/12$.

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$$\frac{\sqrt[]{\cos x}-\sqrt[3]{\cos x}}{\sin^2 x}=\frac{x^4\left(\frac{-24}{x^2}+1\right)}{96x^4\left(\frac{3}{x^2}-1\right)}\Rightarrow L=\frac{-1}{96}$$

This step is incorrect.

The terms within the brackets approach $\infty$ as $x\rightarrow0$, and the limit becomes of the form $\frac{\infty}{\infty}$.

The correct way is

$$\frac{x^4\left(\frac{-24}{x^2}+1\right)}{96x^4\left(\frac{3}{x^2}-1\right)}=\frac{x^2\left(-24+x^2\right)}{96x^2\left(3-x^2\right)}=\frac{\left(-24+x^2\right)}{96\left(3-x^2\right)}$$

$$\lim_{x\rightarrow0}\frac{\left(-24+x^2\right)}{96\left(3-x^2\right)}=\frac{\left(-24\right)}{96\left(3\right)}=\frac{-1}{12}$$

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Hint:

Let $\sqrt[6]{\cos x}=u$

$\sin^2x=1-\cos^2x=1-u^{12}$

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