2
$\begingroup$

I'm having trouble figuring out what to do here, I'm supposed to find this limit:

$$\lim_{x\rightarrow0} \frac{x\cos(x)-\sin(x)}{x^3}$$

But I don't know where to start, any hint would be appreciated, thanks!

$\endgroup$
  • 1
    $\begingroup$ You could use the dreaded Hospital rule, but I would just write down a few terms of the Maclaurin series of the top. $\endgroup$ – Lord Shark the Unknown Nov 21 '17 at 20:21
  • $\begingroup$ I still cannot use L'Hopital rule because we haven't proved that in class. $\endgroup$ – Demian Jimenez Nov 21 '17 at 20:21
  • $\begingroup$ I suppose then that Taylor expansion are forbidden =p $\endgroup$ – Paolo Intuito Nov 21 '17 at 20:27
3
$\begingroup$

Taylor's formula at order $3$ will do (and anyway is better than L'Hospital's rule): \begin{align} \frac{x\cos x-\sin x}{x^3}&=\frac{1}{x^3}\biggl(x\Bigl(1-\frac{x^2}{2}+o(x^2)\Bigr)-\Bigl(x-\frac{x^3}{6}+o(x^3)\Bigr)\biggr)\\ &=\frac{1}{x^3}\biggl(\Bigl(\not x-\frac{x^3}{2}+o(x^3)\Bigr)-\Bigl(\not x-\frac{x^3}{6}+o(x^3)\Bigr)\biggr)\\ &=\frac{1}{x^3}\Bigl(-\frac{x^3}{3}+o(x^3)\Bigr)=-\frac13+o(1)\to -\frac13. \end{align}

$\endgroup$
2
$\begingroup$

Since$$x\cos(x)-\sin(x)=x\left(1-\frac{x^2}2+\cdots\right)-\left(x-\frac{x^3}6+\cdots\right)=-\frac{x^3}3+\cdots,$$your limit is equal to $-\dfrac13$.

$\endgroup$
  • $\begingroup$ Do you think this is more elementary then the L' Hospital rule? $\endgroup$ – Aqua Nov 21 '17 at 20:31
  • $\begingroup$ @JohnWatson It all depends on what one can use to solve the problem and also on how things were defined. If, for the OP,$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\text{ and }\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$$(which I doubt, but it's certainly possible), then my approach is the way to go. $\endgroup$ – José Carlos Santos Nov 21 '17 at 20:37
2
$\begingroup$

One can use L'Hospital's rule or, as shown here, the series expansion of $\sin(x)$ and $\cos(x)$. Using \begin{align} \cos(x) &= 1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} - \cdots \\ \sin(x) &= x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots \end{align} then $$x \cos(x) - \sin(x) = - \frac{2 \, x^{3}}{3!} + \frac{4 \, x^{5}}{5!} - \cdots$$ and $$\lim_{x \to 0} \frac{x \cos(x) - \sin(x)}{x^{3}} = \lim_{x \to 0} \left( - \frac{2}{3!} + \frac{4 \, x^{2}}{5!} - \cdots \right) = - \frac{1}{3}.$$

$\endgroup$
  • $\begingroup$ This is imprecise because of all those ....dots ,not wrong but needs details .Does the poster know about infinite series . A finite Taylor series with remainder might help $\endgroup$ – StuartMN Nov 21 '17 at 20:36
1
$\begingroup$

You could do

Let $u=x^3$

So this gives us

$\lim_{u \rightarrow 0} \frac{ \sqrt[3]{u} \cos(\sqrt[3]{u})-\sin(\sqrt[3]{u})}{ u}$

Let $f(u)=\sqrt[3]{u} \cos(\sqrt[3]{u})-\sin(\sqrt[3]{u})$ And so $f(0)=0$

We know that $\lim_{u \rightarrow 0} \frac{ f(u)-f(0)}{u-0}=\frac{d}{du}(f(u))|_{u=0}$

$\endgroup$
1
$\begingroup$

Hint:

In Equation $(10)$ of this answer, it is shown that $$ \lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag1 $$ Furthermore, as shown in this answer, $\lim\limits_{x\to0}\frac{\sin(x)}x=1$, so $$ \begin{align} \lim_{x\to0}\frac{1-\cos(x)}{x^2} &=\lim_{x\to0}\frac{2\sin^2(x/2)}{4(x/2)^2}\\ &=\frac12\left(\lim_{x\to0}\frac{\sin(x/2)}{x/2}\right)^2\\[3pt] &=\frac12\tag2 \end{align} $$ Limits $(1)$ and $(2)$ can be combined to answer the question.

$\endgroup$
  • $\begingroup$ Both your linked answers are my favorites. I hope OP does have a look at them. $\endgroup$ – Paramanand Singh Nov 22 '17 at 13:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.