# How can I evaluate this limit without using L'hospital, derivatives or series expansion?

I have troubles solving this particular limit without using series expansion, L'Hospital rule or derivatives.

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

I have tried multiplying by $$(\cos{x}+1)$$ and got rid of one $$x$$ in the denominator, but got arguably something even messier. I got this:

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

Any help would be appreciated.

• As $\cos x$ and $\sin x$ are defined as series I think you are asking for the impossible. Mar 25, 2020 at 16:12
• Why do you need to do this without the most suitable tools for the task? Mar 25, 2020 at 16:35
• To piggyback off of the comments above, any method that doesn't involve series will inevitably require a squeeze theorem argument that is more or less derived from series. Mar 25, 2020 at 16:40
• @ClementC.: for me this does give a thrill while tackling difficult problems using simpler tools. Occasionally one should enjoy such problems but not make them routine. Mar 27, 2020 at 8:56
• @ParamanandSingh Sometimes, the solution using simple tools is elegant. Sometimes, it just ends up long, technical, and contrived... Mar 27, 2020 at 11:13

Various limits are proved in this answer using basic tools only. In particular, we need $$\lim_{t\to 0}\sin(t)/t=0$$, $$\lim_{t\to 0}(\cos(t)-1)/t=0$$, and $$\lim_{t\to 0}(t-\sin(t))/t^3=1/6$$. \begin{align} \frac{x\sin x+2(\cos x-1)}{x^3}&=\frac{x(2\sin(x/2)\cos(x/2))+2(-2\sin^2(x/2))}{x^3} \\ &= \frac{\sin(x/2)}{x/2}\cdot\frac{x\cos(x/2)-2\sin(x/2)}{x^2} \end{align} We have $$\lim_{x\to 0}\sin(x/2)/(x/2)=1$$. As for the second factor, \begin{align} \frac{x\cos(x/2)-2\sin(x/2)}{x^2} &= \frac{\left[x(\cos(x/2)-1)\right]+\left[\sin(x)-2\sin(x/2)\right]+\left[x-\sin(x)\right]}{x^2} \\ &= \frac{\cos(x/2)-1}{x}+\frac{2\sin(x/2)(\cos(x/2)-1)}{x^2}+\frac{x-\sin(x)}{x^2} \\ &= \frac 12\frac{\cos(x/2)-1}{x/2}+\frac 12\frac{\sin(x/2)}{x/2}\frac{\cos(x/2)-1}{x/2}+\frac{x-\sin(x)}{x^3}\cdot x \end{align} You can finish with the aforementioned limits. The limit of the second factor is $$0$$. So the desired limit is $$1\cdot 0=0$$.

• Brilliant, that's exactly what I was hoping for!! Mar 25, 2020 at 17:40

I'll make use of some limits famously provable without the banned techniques:\begin{align}\frac{x\sin x+2(\cos x-1)}{x^3}&=\frac{2x\sin\frac{x}{2}\cos\frac{x}{2}-4\sin^2\frac{x}{2}}{x^3}\\&=\underbrace{\frac{4\sin^2\frac{x}{2}}{x^2}}_{\to1}\underbrace{\frac{\frac{x}{2}\cot\frac{x}{2}-1}{x}}_{\to0}.\end{align}The proof that $$\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\tan x}{x}=1$$ is famous, whence$$\lim_{x\to0}\frac{x\cot x-1}{x}=\lim_{x\to0}\frac{x-\tan x}{x\tan x}=\lim_{x\to0}\frac{x-\tan x}{x^2}=0$$because $$\lim_{x\to0}\frac{x-\tan x}{x^3}=-\frac13$$ is famous too (see e.g. @bjorn93's link above).

• One can prove that $(x-\tan x) /x^2\to 0$ using the famous inequality $\sin x <x<\tan x$ combined with Squeeze. Mar 27, 2020 at 8:52

Whether series is allowed or not, here's the answer.

$$x\sin x = x(x-x^3/3!+x^5/5!...)$$ so the lowest order term is $$x^2$$. Simiilarly,

$$2(1-\cos x)=2[1-(1-x^2/2!+x^4/4!...)]=2[x^2/2-x^4/4!...]$$ so the lowest order term is $$x^2$$.

When you subtract the second from the first, the $$x^2$$ terms will cancel. The next lowest order in the numerator is $$O(x^4)$$. Then it is clear that the limit is $$0$$.

Note,

$$\frac{x\sin x + 2(\cos x -1)}{x^3} =\frac{\sin x}{x}\cdot \frac {x-2\tan\frac x2}{x^2}$$ and,

$$x-2\tan\frac x2 = O(x^3)$$

Thus,

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

Let $$x\in\mathbb{R}^{*}$$, observe that : $$\fbox{\begin{array}{rcl}\displaystyle\frac{x-\sin{x}}{x^{2}}=\frac{x}{2}\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\end{array}}$$

Using the fact that $$\left(\forall t\in\left[0,1\right]\right),\ \left|\cos{\left(tx\right)}\right|\leq 1$$, we have : \begin{aligned} \left|\frac{x-\sin{x}}{x^{2}}\right|=\frac{\left|x\right|}{2}\left|\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\right|&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\left|\cos{\left(tx\right)}\right|\mathrm{d}t}\\&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\,\mathrm{d}t} \end{aligned}

Which means $$\left(\forall x\in\mathbb{R}^{*}\right),\ \left|\frac{x-\sin{x}}{x^{3}}\right|\leq\frac{\left|x\right|}{6}$$, and thus $$\lim\limits_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 \cdot$$

Let $$x\in\mathbb{R}^{*}$$, observe that : $$\fbox{\begin{array}{rcl}\displaystyle\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}=\int_{0}^{1}{\left(1-t\right)^{2}\sin{\left(tx\right)}\,\mathrm{d}t}\end{array}}$$

Using the fact that $$\left(\forall t\in\left[0,1\right]\right),\ \left|\sin{\left(tx\right)}\right|\leq t\left|x\right|$$, we have : \begin{aligned} \left|\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}\right|=\left|\int_{0}^{1}{\left(1-t\right)^{2}\sin{\left(tx\right)}\,\mathrm{d}t}\right|&\leq\int_{0}^{1}{\left(1-t\right)^{2}\left|\sin{\left(tx\right)}\right|\mathrm{d}t}\\&\leq\left|x\right|\int_{0}^{1}{t\left(1-t\right)^{2}\,\mathrm{d}t} \end{aligned}

Which means $$\left(\forall x\in\mathbb{R}^{*}\right),\ \left|\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}\right|\leq\frac{\left|x\right|}{12}$$, and thus $$\lim\limits_{x\to 0}{\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}}=0 \cdot$$

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

As J.G. already shown it holds \begin{align}\frac{x\sin x+2(\cos x-1)}{x^3}&=\frac{2x\sin\frac{x}{2}\cos\frac{x}{2}-4\sin^2\frac{x}{2}}{x^3}\\&=\frac{4\sin^2\frac{x}{2}}{x^2}\frac{\frac{x}{2}\cot\frac{x}{2}-1}{x} \\ &= \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \frac{f(x) - f(0)}{x - 0} \end{align}

for $$f(x) = x\cot(x)$$

So we get by the mean value theorem: $$\lim_{x\to 0} \frac{x\sin x+2(\cos x-1)}{x^3} = 0\cdot\lim_{x\to 0} f'(x) = 0\cdot1 = 0$$