How to solve the limit of the 2nd derivative of $\frac{\sin(x)}{x}$ as $x \to 0$

Question

Given the function $\frac{(x^2-2)\sin(x)+2x\cos(x)}{x^3}$, I would like to find the limit as $x\to0$. The following is my approach:

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

Hence, I theoretically obtain DNE as my answer. However, when I plotted the function, the limit is $\frac{1}{3}$. I would like to get some advice on how should I get around this problem?

Are you allowed to use l'Hôpital's rule and/or Taylor series? — StackTD, Apr 13 '17 at 12:44
Nope. Yet to learn L'Hopital. Is it possible to solve it without using those? — M.A.N, Apr 13 '17 at 12:46
If you can't use those, you will need something such as one or more standard limits you have seen and are allowed to use? E.g. $\tfrac{\sin x}{x} \to 1$ for $x \to 0$? — StackTD, Apr 13 '17 at 12:48
@M.A.N See comment above; are there (standard) limits you are allowed to use? If so, which one(s)? You'll need some machinery. — StackTD, Apr 13 '17 at 12:57
@StackTD: Yeap, I get $\frac{sin(x)}{x} \to 1 for x \to 1$ but I still can't get the idea to solve it because the second and third terms just burst to infinity. — M.A.N, Apr 13 '17 at 12:58
$(x^2-2)\sin x + 2x\cos x = x^2\sin x + 2(x - \sin x) + 2x(\cos x - 1)$ — Daniel Fischer, Apr 13 '17 at 12:59
You can mark one of the answers as accepted (which is probably better than editing the title to say "solved"). See here for more details. — Martin Sleziak, Apr 13 '17 at 13:49

score 3 · Accepted Answer · answered Apr 13 '17 at 13:04

After comments, you can't use l'Hôpital or Taylor series but you can use the well-known limit: $$\lim_{x \to 0} \frac{\sin x}{x} = 1 \tag{$\star$}$$ Rewrite: $$\frac{(x^2-2)\sin x+2x\cos x}{x^3} = \color{blue}{\frac{x^2}{x^2}\frac{\sin x}{x}}+2\color{red}{\frac{x\cos x-\sin x}{x^3}}$$ Taking the limit gives $1$ for the blue part and for the red part you have: $$\lim_{x \to 0} \frac{x\cos x-\sin x}{x^3} = -\frac{1}{3}$$ which is shown in this answer using only $(\star)$; together: $$\lim_{x \to 0}\frac{(x^2-2)\sin x+2x\cos x}{x^3} =\lim_{x \to 0} \left( \color{blue}{\frac{x^2}{x^2}\frac{\sin x}{x}}+2\color{red}{\frac{x\cos x-\sin x}{x^3}} \right) = \color{blue}{1}+2\left(\color{red}{-\tfrac{1}{3}}\right) = \tfrac{1}{3}$$

@M.A.N You're welcome. Rewriting a bit differently (see also Daniel Fischer♦'s comment) can lead to other well-known limits, but it will always boil down to using the $(\star)$-one to derive the other ones and then properly combining them. — StackTD, Apr 13 '17 at 13:18

score 2 · Answer 2 · edited Apr 13 '17 at 13:41

2

You can separate first term and keep remaining terms as same this yields you a function as $1-\left(\lim_{x\to 0}\frac {2\sin(x)+2x\cos(x)}{x^3}\right)$. Now use l'Hospital's rule and the fact that: $$\lim_{x\to 0}\frac {\sin(x)}{x}=1\,.$$

edited Apr 13 '17 at 13:41

answered Apr 13 '17 at 12:49

Archis Welankar

15,910

score 2 · Answer 3 · answered Apr 13 '17 at 12:54

Use Τaylor's expansion at order $3$: the numerator is $$(x^2-2)\Bigl(x-\frac{x^3}6+o\bigl(x^3\bigr)\Bigr)+2x\Bigl(1-\frac{x^2}2+o\bigl(x^3\bigr)\Bigr)=\frac{x^3}3+o\bigl(x^3\bigr)\sim_0\frac{x^3}3,$$ so $\qquad\dfrac{(x^2-2)\sin(x)+2x\cos(x)}{x^3}\sim_0\dfrac{x^3}{3x^3}=\dfrac 13.$

How to solve the limit of the 2nd derivative of $\frac{\sin(x)}{x}$ as $x \to 0$

3 Answers3