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.
Best Answer
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 $1\cdot 0=0$.