Finding $\lim_{x\to 0}\frac{\sin(x+x^3/6)-x}{x^5}$

Question

I'm trying to find the limit of this expression:

$$\lim_{x\to0}\frac{\sin\left(x+x^3/6\right)-x}{x^5}$$

My solution is as follows: $$ \begin{align} \lim_{x\to0}\frac{\sin\left(x+x^3/6\right)-x}{x^5}&=\lim_{x\to0}\frac{1}{x^5}\cdot\!\!\left[\frac{(x+x^3/6)\sin(x+x^3/6)}{(x+x^3/6)}-x\right]\\ &=\lim_{x\to0}\frac{1}{x^5}\cdot\!\!\left[(x+x^3/6)-x\right]\\ &=\lim_{x\to0}\frac{x^3}{6x^5}\\ &=\lim_{x\to0}\frac{1}{6x^2}=+\infty \end{align} $$ But Wolfram Alpha finds the limit as $-\frac{3}{40}$. Where is my mistake?

Answer 1

For the error, see the comments and other answers which address it already. If you are familiar with Taylor series, you can compute your limit as follows, observing that $$ \sin y \operatorname*{=}_{y\to 0} y - \frac{y^3}{6} + \frac{y^5}{120} +o(y^5)\ . $$ $$\begin{align} \sin (x+\frac{x^3}{6}) &\operatorname*{=}_{x\to 0} x+\frac{x^3}{6} - \frac{(x+\frac{x^3}{6})^3}{6} + \frac{(x+\frac{x^3}{6}^5)}{120} +o(x^5) \\ &= x+\frac{x^3}{6} -\left(\frac{x^3}{6}+\frac{x^5}{12}\right) + \frac{x^5}{120 } +o(x^5) \\ &= x-\frac{3}{40}x^5 +o(x^5) \end{align}$$ where we ignore terms $x^k$ for $k > 5$ (as they get "swallowed" in the $o(x^5)$) when developing the $(\cdot)^3$ and $(\cdot)^5$. Note that we need to go at least to the second term of the sinus (in $y^3$) as the first will be canceled out by the $-x$ that we are about to add: $$\begin{align} \frac{\sin (x+\frac{x^3}{6}) - x}{x^5}&\operatorname*{=}_{x\to 0} \frac{x-\frac{3}{40}x^5 +o(x^5) - x}{x^5} = \frac{-\frac{3}{40}x^5 +o(x^5)}{x^5} = -\frac{3}{40} + o(1) \end{align}$$ yielding the limit.

Answer 2

Your mistake is in this line

$$\lim_{x\to0}\dfrac{\left[{\left(x+\tfrac{x^3}6\right)\sin\left(x+\tfrac{x^3}6\right)}\left/\right.{\left(x+\tfrac{x^3}6\right)}\right]-x}{x^5}=\lim_{x\to0}\dfrac{x+\tfrac16x^3-x}{x^5}.$$

You can't evaluate the limit using $\lim\limits_{u\to0}\frac{\sin u}u=1$ directly since you're using the two laws $$\lim_{x\to0}(f+g)=\lim_{x\to0}f+\lim_{x\to0}g,\quad\lim_{x\to0}f\cdot g=\lim_{x\to0}f\cdot\lim_{x\to0} g,$$ and they don't hold when $\lim f$ or $\lim g$ don't exist. (or both)

Answer 3

This limit can be calculated by applying (fiercely :)) L'Hospital's rule three times in succession: $$\lim_{x\to0}\frac{\sin\left(x+x^3/6\right)-x}{x^5}=$$I.$$=\lim_{x\to0}\frac{(1+x^2/2)\cos\left(x+x^3/6\right)-1}{5x^4}=$$ II.$$=\lim_{x\to0}\frac{x\cos\left(x+x^3/6\right)-(1+x^2/2)^2\sin\left(x+x^3/6\right)}{20x^3}=$$ III.$$=\lim_{x\to0}[\frac{\cos\left(x+x^3/6\right)-x(1+x^2/2)\sin\left(x+x^3/6\right)}{60x^2}-$$$$-\frac{2x(1+x^2/2)\sin\left(x+x^3/6\right)+(1+x^2/2)^3\cos\left(x+x^3/6\right)}{60x^2}]=$$ $$=\frac{1}{60}\lim_{x\to0}[\frac{\cos\left(x+x^3/6\right)(1-(1+x^2/2)^3)}{x^2}-\frac{3x(1+x^2/2)\sin\left(x+x^3/6\right)}{x^2}]=$$$$= \frac{1}{60}(-\frac{3}{2}-3) =-\frac{3}{40}.$$