What is the result of $\lim\limits_{x \to 0}(1/x - 1/\sin x)$?

For fun, and because of the pre-calculus tag, we give a proof without calculus. It turns out that there is a geometric argument that $|x-\sin x|$ is less than a constant times $|x^3|$ for $x$ near $0$.

I will need some help from you, to draw the missing picture. We have $$\frac{1}{x}-\frac{1}{\sin x}=\frac{\sin x-x}{x\sin x}.$$ Let
$$f(x)=\frac{x-\sin x}{x\sin x}$$ (the change of sign is for convenience). We will show that $\lim\limits_{x\to 0}\,f(x)=0.$

We are interested in the behaviour of $f(x)$ when $x$ is close (but not equal) to $0$. Note that $f(-x)=-f(x)$. So we will be finished if we can show that $f(x)$ approaches $0$ as $x$ approaches $0$ through positive values.

Let $x$ be small positive. Draw $\triangle OPQ$ as follows. The base of the triangle is $OP$, and has length $1$. The triangle is right-angled at $P$. Finally, $Q$ is such that $\angle QOP =x$.

Draw the circular sector with centre $O$, radius $1$, and going from $P$ to a point on $OQ$. So the sector has angle $x$.

Note that the circular sector is contained in $\triangle OPQ$. The circular sector has area $(1/2)x$, and $\triangle OPQ$ has area $(1/2)\tan x$. Thus the geometry gives us the inequality $$\frac{x}{2}<\frac{\tan x}{2}.$$ Since $x>\sin x$, we get the estimates $$0<x-\sin x< \tan x-\sin x.$$ The right-hand side only involves trigonometric functions, so is easier to deal with than $x-\sin x$: $$\tan x-\sin x=\sin x\left(\frac{1-\cos x}{\cos x}\right)=\sin x\left(\frac{1-\cos^2 x}{\cos x(1+\cos x)}\right)=\frac{\sin^3 x}{\cos x(1+\cos x)}.$$ We conclude that $$0 <\frac{x-\sin x}{x\sin x}<\frac{\sin^2 x}{x\cos x(1+\cos x)}.$$ Since $\sin x<x$, we find that $$0 <\frac{x-\sin x}{x\sin x}<\frac{\sin x}{\cos x(1+\cos x)},$$ and it is clear that $\dfrac{\sin x}{\cos x(1+\cos x)}$ approaches $0$ as $x$ approaches $0$ through positive values.

Comment: In this problem, there is no virtue in avoiding the calculus. The Taylor expansion is the natural approach.

Hint: Try using $$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\left(\frac{\sin x - x}{x\sin x}\right)$$ and apply L'Hopital's rule.

Simplify to have $$\frac{\sin x-x }{x\sin x}$$ and consider Maclaurin's series for $$\sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}-...$$

So you have $$\frac{(x-\frac {x^3}{3!}+\frac {x^5}{5!}-...)-x}{x(x-\frac {x^3}{3!}+\frac {x^5}{5!}+...)}=\frac{(-\frac {x}{3!}+\frac {x^3}{5!}-...)}{(1-\frac {x^2}{3!}+\frac {x^4}{5!}-...)}.$$

Finding the limit as $x\rightarrow 0$, we have;

$$\frac{\lim_{x\rightarrow 0}(-\frac {x}{3!}+\frac {x^3}{5!}-...)}{\lim_{x\rightarrow 0}(1-\frac {x^2}{3!}+\frac {x^4}{5!}-...)}=\frac{0}{1}=0.$$

which is the required answer.