Does $\lim_{x\to 0}(\frac{1}{x}-\frac{\sin x}{x^2})$ exist? $$\lim_{x\to0}\left(\frac{1}{x}-\frac{\sin x}{x^2}\right)=\lim_{x\to0}\left(\frac{x-\sin x}{x^2}\right)$$
By L'Hopital's Rule, we have
$$\lim_{x\to0}\left(\frac{x-\sin x}{x^2}\right)=\lim_{x\to0}\frac{1-\cos x}{2x}=\lim_{x\to0}\frac{\sin x}{2}=0$$
This is what I did, but the answer in my textbook says that the limit does not exist. Can someone explain to me where have I done it wrong? Thx!
 A: As others have pointed out, the OP's answer is correct; the limit is amenable to L'Hopital, and equals $0$. It's worth noting, though, that one can avoid L'Hopital, using instead the inequality $\sin x\ge x-{1\over3}x^3$ for $x\ge0$. This inequality and the symmetry $\sin(-x)=-\sin x$ give us
$$|x-\sin x|\lt{1\over3}|x|^3$$
for all $x$, hence
$$\left|{1\over x}-{\sin x\over x^2}\right|\lt{1\over3}|x|\to0$$
as $x\to0$.
A: Yes, there exists the limit of $\frac{1}{x}-\frac{\sin(x)}{x^{2}}$ as $x\to 0$ or  as $x\to \infty$ and the both case the answer in zero.
Your solution:
\begin{eqnarray*}
\lim_{x\to 0} \left(\frac{1}{x}-\frac{\sin(x)}{x^{2}}\right)&=&\lim_{x \to 0} \left(\frac{x-\sin(x)}{x^{2}} \right) \quad [\color{blue}{\text{correct}}]\\
&=&\lim_{x\to0}\frac{1-\cos x}{2x} \quad [\color{blue}{\text{correct}}]\\
&=&\lim_{x\to0}\frac{\sin x}{2} \quad [\color{blue}{\text{correct}}]\\
&=&0. \quad [\color{blue}{\text{correct}}]\\
\end{eqnarray*}

On the other hand, notice that in $$\lim_{x\to \infty} \left(\frac{1}{x}-\frac{\sin(x)}{x^{2}}\right)$$
we can see that $$\lim_{x\to \infty} \frac{1}{x^{2}}=0 \quad \text{and} \quad \lim_{x\to \infty} \frac{\sin(x)}{x^{2}}=0$$
since that $-1\leq \sin(x) \leq 1$ for all $x\in  \mathbb{R}$.
Therefore, we can conclude that $$\lim_{x\to \infty} \left(\frac{1}{x}-\frac{\sin(x)}{x^{2}}\right)=0.$$
A: To explore whether the limit exists or not proceed with calculating the Left Hand Limit and the Right Hand Limit.
Say, we approach the limit from values slightly greater than $0$ i.e. $x$ approaches $0^+$ where $0^+$ denotes values that are slightly greater than $0$. This is known as the Right Hand Limit (RHL):
$$\lim_{x\to0^+}\left(\frac{1}{x}-\frac{\sin x}{x^2}\right)=\lim_{x\to0^+}\left(\frac{x-\sin x}{x^2}\right)$$
Since, $x\to0^+$, we can use the substitution $x=0+h$ ($x$ approaches $0$ from slightly higher value). As, $x\to0$, $h\to0$. So the problem becomes:
$$\lim_{h\to0}\left(\frac{1}{h}-\frac{\sin h}{h^2}\right)=\lim_{h\to0}\left(\frac{h-\sin h}{h^2}\right)$$
$$\lim_{h\to0}\frac{h-(h-\frac{h^3}{3!}+\frac{h^5}{5!}...)}{h^2}=\lim_{h\to0}{\frac{\frac{h^3}{3!}-\frac{h^5}{5!}+\frac{h^7}{7!}}{h^2}}=0$$
Similarly, you can calculate the Left Hand Limit by assuming, $x\to0^-$ ($x$ approaches $0$ from values slighltly less than $0$). The LHL becomes:
$$\lim_{x\to0^-}\left(\frac{1}{x}-\frac{\sin x}{x^2}\right)=\lim_{x\to0^-}\left(\frac{x-\sin x}{x^2}\right)$$
Use substitution, $x=0-h$. So, as $x\to0^-$,$h\to0$. The LHL becomes:
$$\lim_{h\to0}\left(\frac{1}{-h}-\frac{\sin (-h)}{(-h)^2}\right)=\lim_{h\to0}\left(\frac{-h+\sin h}{h^2}\right)$$
$$\lim_{h\to0}\frac{-h+(h-\frac{h^3}{3!}+\frac{h^5}{5!}-\frac{h^7}{7!})}{h^2}=\lim_{h\to0}\frac{\frac{-h^3}{3!}+\frac{h^5}{5!}...}{h^2}=0$$
You can see that the LHL = RHL, therefore the limit exists and is equal to $0$. Your method is correct as well but this method always let you see whether the limits exist or not. I hope this helps...
