Solve without using L'Hôpital's rule? 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.

Let $\displaystyle L=\lim_{x \to 0} \frac{\sin x - x}{x^2}$. Replacing $x$ by $2y$, we obtain:
$
\begin{align*}
L & = \lim_{y \to 0} \frac{\sin 2y - 2y}{4y^2} = \lim_{y \to 0} \frac{2 \sin y \cos y - 2y}{4y^2} = \lim_{y \to 0} \frac{2 \sin y \cos y - 2 \sin y + 2 \sin y - 2y}{4y^2} \\
& = \lim_{y \to 0} \left( \frac{2 \sin y(\cos y-1)}{4y^2}+\frac{1}{2} \lim_{y \to 0} \frac{\sin y - y}{y^2}\right) \\
& = \frac{1}{2} \lim_{y \to 0} \sin y \cfrac{-2\sin^2\cfrac{y}{2}}{y^2}+\frac{1}{2}L \\ 
& = -\frac{1}{4} \lim_{y \to 0} \sin y \cfrac{\sin^2\cfrac{y}{2}}{\cfrac{y^2}{4}}+\frac{1}{2}L, 
\end{align*}
$
which implies that $L=0$.
Hence,
$
\begin{align*}
\lim_{x\to 0} \left(\frac{3x+1}{x}-\frac{1}{\sin x}\right) & = \lim_{x\to 0} \left(3+\frac{1}{x}-\frac{1}{\sin x}\right) = \lim_{x\to 0} \left(3+\frac{\sin x - x}{x \sin x}\right) \\
& = \lim_{x\to 0} \left(3+\frac{\sin x - x}{x^2} \cdot \frac{x}{\sin x} \right) = 3+0 \cdot 1 \\ 
& = \boxed{3},
\end{align*}
$
