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.
Via power series

$$\frac{3x+1}{x}-\frac{1}{\sin x}=\frac{3x\sin x+\sin x-x}{x\sin x}=$$

$$\frac{3x\left(x-O(x^3)\right)+\left(x-O(x^3)\right)-x}{x\left(x-O(x^3)\right)}=\frac{3x^2-O(x^4)-O(x^3)}{x^2-O(x^4)}=$$

$$3\frac{1-O(x^2)-O(x)}{1-O(x^2)}\xrightarrow [x\to 0]{}3\cdot 1=3$$
alright so lets break up the first term to get

lim(x->0) [3x/x + 1/x - 1/sinx] 
now lets use properties of limits 

lim(x->0) 3 + lim(x->0) (1/x) - lim(x->0) 1/sinx
now knowing that lim(x->0) sinx/x = 1 lets divide the third term by x at the numerator as well as the denominator to obtain

lim(x->0) 3 + lim(x->0) (1/x) - lim(x->0) (1/x)/sinx/x

and since the denominator is just 1 we are left with

lim(x->0) 3 + lim(x->0) (1/x) - lim(x->0) (1/x)

BUT notice lim x--> 0 1/x does not exist. so this direct method of finding limit would not work. another use is power series as one of the answer is posted :)