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For evaluating the limit $\lim\limits_{x \to 0} \frac{x(1 - 0.5\cos x) - 0.5\sin x}{x^3}$, I proceeded as follows:

$$\lim_{x \to 0} \left(\frac{x(1 - 0.5\cos x)}{x^3} - \frac{0.5}{x^2}\left(\frac{\sin x}{x}\right)\right)$$

Using the fact that $\lim\limits_{x \to 0} \frac{\sin x}{x}=1$ and $\lim\limits_{x \to a} \{ f(x) - g(x)\} = \lim\limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$, I got $$\lim\limits_{x \to 0} \frac{1 - 0.5\cos x}{x^2} - \lim\limits_{x \to 0} \frac{0.5}{x^2}$$ $$ = \lim_{x \to 0} \left(\frac{1 - 0.5\cos x - 0.5}{x^2} \right)$$

Now, after applying L'Hopital's Rule I got the final answer as $0.25$. However, on evaluating the original limit using Mathematica, I got the answer as $\frac{1}{3}$.

Can someone please tell me where am I going wrong.

Somehow, I believe that you cannot use the fact that $\lim\limits_{x \to a} \{ f(x) - g(x)\} = \lim \limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$ in case of indeterminate forms.

Thanks in advance!

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    $\begingroup$ The first problem is that you can't break up the limit like this. In particular, you can only break up the limit into two limits when both of the limits exist. Since $\lim_{x\rightarrow 0}\frac{0.5}{x^2}$ does not exist (and neither does $\lim_{x\rightarrow 0}\frac{1-0.5\cos(x)}{x^2}$), you would end up with $\infty-\infty$, which is indeterminate. $\endgroup$ – Michael Burr Nov 22 '15 at 15:08
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    $\begingroup$ Hint: Just apply L'Hopital to the original expression, there is no need to break it up into smaller expressions (especially since that introduces an error). $\endgroup$ – Michael Burr Nov 22 '15 at 15:09
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    $\begingroup$ Apply L'Hospital three times successively, and you get 2/6 which is 1/3. $\endgroup$ – barrycarter Nov 22 '15 at 15:11
  • $\begingroup$ Whoever writes the software used by those who use software to write MathJax code rather than writing MathJax code by hand must be having a competition to see who can make the code look more psychotic than anyone else's. I cleaned this one up. $\endgroup$ – Michael Hardy Nov 22 '15 at 16:46
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If you want to compute this limit by making use of known basic limits then use the following \begin{equation*} \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}=-\frac{1}{6},\ \ \ \ and\ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2} \end{equation*} as follows \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{x(1-\frac{1}{2}\cos x)-\frac{1}{2}\sin x}{x^{3}} &=&\lim_{x\rightarrow 0}\frac{1}{2}\frac{2x(1-\frac{1}{2}\cos x)-\sin x}{% x^{3}} \\ &=&\lim_{x\rightarrow 0}\frac{1}{2}\frac{x+x-x\cos x-\sin x}{x^{3}} \\ &=&\lim_{x\rightarrow 0}\frac{1}{2}\left( \frac{x-\sin x}{x^{3}}+\frac{% 1-\cos x}{x^{2}}\right) \\ &=&\frac{1}{2}\left( \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}% +\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}\right) \\ &=&\frac{1}{2}\left( -\frac{1}{6}+\frac{1}{2}\right) =\frac{1}{6}. \end{eqnarray*}

Warning: When we write \begin{equation*} \lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}+\frac{1-\cos x}{x^{2}}% \right) =\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}\right) +\lim_{x\rightarrow 0}\left( \frac{1-\cos x}{x^{2}}\right) \end{equation*} it is because we $\textbf{previously}$ know that the limits of the RHS exist as real numbers.

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Use Taylor's development at order $3$:

  • $\cos x= 1-\dfrac{x^2}2+o(x^2)$, hence $\;x\bigl(1-\frac12\cos x\big)=\dfrac x2+\dfrac{x^3}4+o(x^3)$,
  • $\sin x=x-\dfrac{x^3}6+o(x^3)$.

Thus the numerator is $$\dfrac x2+\dfrac{x^3}4-\frac x2+\dfrac{x^3}{12}+o(x^3)=\dfrac{x^3}3+o(x^3)$$ and finally $$\frac{x\bigl(1-\frac12\cos x\big)-\frac12\sin x}{x^3}=\frac{\dfrac{x^3}3+o(x^3)}{x^3}=\frac13+o(1)\to\frac13.$$

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Apply iteratedly l'Hôpital Rule to $$\frac{x(2-\cos x)-\sin x}{2x^3}$$ We get successively $$\frac{2-2\cos x+x\sin x}{6x^2}$$ $$\frac{3\sin x+x\cos x}{12x}$$ $$\frac{3\cos x+\cos x-x\sin x}{12}$$

Thus the searched limit is the limit of $$\frac{4\cos x-x\sin x}{12}$$ when $x$ tends to $0$ which is clearly $\frac{4}{12}=\frac 13$

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  • $\begingroup$ @Michael Hardy: My English is very weak (just for reading and zero for speaking) and I can not see what you have edited. Thanks anyway. You have a very prestigious last name in mathematics (by G. H. Hardy) $\endgroup$ – Piquito Nov 23 '15 at 11:01
  • $\begingroup$ If you write 3 sin x in MathJax, you see $3 sin x$, but if you write 3\sin x, with a backslash, you see $3\sin x$, with proper spacing before and after $\sin$ and with $\sin$ not italicized. If you write $3\sin(x)$, coded as 3\sin(x), then you don't have the same amount of space after $\sin$ as when you write $3\sin x$. My edit changed only some of the MathJax code. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 23 '15 at 18:49

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