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I'm considering this limit $$\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}.$$ My first attempt was using the following equivalent infinitesimals $$1-\cos x^2 \sim \frac{x^4}{2},\quad \arctan x \sim x, \quad \sin 2x \sim 2x, \quad e^{x^2} - 1 \sim x^2 \,\,\,\,\text{when}\,\,\,\,x\rightarrow 0,$$ and then \begin{align*} \lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2xe^{x^2}-2x+\frac{2}{3}x^3}=\lim_{x\to 0}\frac{x^4}{4(e^{x^2}-1)+\frac{4}{3}x^2}\\ &=\lim_{x\to 0}\frac{x^4}{4x^2+\frac{4}{3}x^2}=\lim_{x\to 0}\frac{3}{16}x^2=0. \end{align*} Later, inspecting this other approach combining infinitesimals and the Taylor expansions \begin{align*} e^{x^2} &= 1+x^2+\frac{x^4}{2}+o(x^4),\\ \sin 2x &= 2x - \frac{4x^3}{3} + \frac{4x^5}{15} + o(x^5),\\ e^{x^2}\sin 2x &= 2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5), \end{align*} I get this other result \begin{align*} \lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5)-2x+\frac{2}{3}x^3}\\ &=\lim_{x\to 0}\frac{x^5/2}{-x^5/15 + o(x^5)}=-\frac{15}{2}. \end{align*} Can someone help me to identify what am I getting wrong?

Thanks.

Gustavo
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  • In the first, the method is incorrect since you do not have the right to replace a subexpression with its equivalent. As @JoséCarlosSantos pointed out, this amounts to making equivalent sums and this is only possible under certain specific conditions. But you're lucky the result is good! By the way, by doing this, you ignore for example the term $-\frac{4x^3}{3}$ in the development of $\sin (2x)$. – Jean-Claude Colette Jan 23 '20 at 11:19
  • This is a basic misconception that one can replace $A$ by $B$ in calculus even if $A\neq B$. One can't replace $A$ by $B$ unless $A=B$. If you see such invalid replacements working fine then it is a case of either pure luck or some detailed step missing. – Paramanand Singh Jan 23 '20 at 11:48

3 Answers3

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Your first method is correct and the answer is indeed $0$. The problem with your second approach is that the Taylor series of $e^{x^2}\sin(2x)$ centered at $0$ begins with $2x\color{red}+\frac23x^3$.

José Carlos Santos
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Keep in mind that you cannot add equivalents inconsiderately. So you second approach is correct. A simpler counter example is

$$f(x)={\sin{x}-x\over x^2}$$

If I just take equivalents I find $f(x)\to \infty$ obviously incorrect because the limit is $0$ as the Taylor expansions show.

marwalix
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Both computations are incorrect. The first method amounts to writing \begin{equation} \begin{array}{l} 1 - \cos(x^2) = \frac{x^4}{2} + o(x^4)\cr \arctan(x) = x + o(x)\cr \sin(2x)= 2x + o(x)\cr e^{x^2}= 1 + x^2 + o(x^2) \end{array} \end{equation} but substituting these values in the fraction cannot tell us the limit because it gives \begin{equation} \frac{\frac{x^5}{5}+ o(x^5)}{o(x)} \end{equation} which limit is not known.

In the second method, as @JoséCarlosSantos pointed out, there is a wrong sign. It is most probable that the sign before $\frac{2}{3}x^3$ is already wrong in the initial fraction. Apart from this point, the second method should work.

Gribouillis
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