TITLE: Solve without using L'Hôpital's rule?
QUESTION [9 upvotes]: Is it possible to solve this without using L'Hôpital's rule?
$$
\lim_{x\to 0} \Big(\frac{3x+1}{x}-\frac{1}{\sin x}\Big)
$$
I tried to solve it but I got stuck at the $\frac{1}{\sin x}$ part.
$$
=\lim_{x\to 0}\frac{3+x^{-1}}{1}-\lim_{x\to 0}\frac{1}{\sin x}
$$
since $\lim_{x\to 0}\frac{1}{\sin x}$  does not exist.

REPLY [4 votes]: Here is a hint which uses no L'Hôpital, no derivatives, and no power series; just trigonometry.
Hint: In this answer, it is shown, geometrically, that $\sin(x)\le x\le\tan(x)$ for $x\in[0,\pi/2)$. Thus,
$$
\begin{align}
\left|\,\frac{1}{x}-\frac{1}{\sin(x)}\,\right|
&=\left|\,\frac{\sin(x)-x}{x\sin(x)}\,\right|\\
&\le\left|\,\frac{\sin(x)-\tan(x)}{x\sin(x)}\,\right|\\
&=\left|\,\frac{1-\sec(x)}{x}\,\right|\\
&=\left|\,\frac{1-\sec(x)}{x}\frac{1+\sec(x)}{1+\sec(x)}\,\right|\\
&=\left|\,\frac{1-\sec^2(x)}{x(1+\sec(x))}\,\right|\\
&=\left|\,\frac{-\tan^2(x)}{x(1+\sec(x))}\,\right|\\
&=\left|\,\frac{\tan(x)}{x}\frac{\tan(x)}{1+\sec(x)}\,\right|\\
&=\left|\,\frac{\tan(x)}{x}\tan(x/2)\,\right|
\end{align}
$$