Limit as $x$ approaching $0$ of$ (9/x) - 9cot(x)$ Ok so the correct answer is $0$ and it is confirmed graphically, but how do we conclude this algebraically? This is breaking the property of limits where u can take the individual limits of $(9/x)$ and $9cotx$ and the limit of the function would be $L - C$, but $L$ and $c$ does not exist? 
 A: $$\lim_{x\to 0} \frac9x - \frac{9\cos x}{\sin x} = \lim _{x\to 0}  \frac{9\sin x - 9x\cos x}{x\sin x}$$
Finding a common denominator for the two fractions, we obtain a limit of the form $\frac00$. Now, we can apply L’Hospital’s rule. It takes two applications to get an answer:
$$\begin{align}
\lim _{x\to 0}  \frac{9\sin x - 9x\cos x}{x\sin x} &= 9\lim _{x\to 0}  \frac{x\sin x}{x\cos x+\sin x}\\
&=9\lim _{x\to 0}  \frac{x\cos x + \sin x}{2\cos x-x\sin x}
\end{align}$$
This can also be done with Laurent series (or Taylor series, or Fourier series), but that’s a bit heavier, theory-wise.
A: Just take into account that for small $x$, $\cot x \approx 1/x$. Otherwise, split the limits and consider the Fourier series of $\cot x$, which is
$$
\cot x = \frac{1}{x} + \sum_{n = 1}^\infty \frac{(-1)^n 2^{2n}B_{2n}}{(2n)!}x^{2n - 1}.
$$ 
A: Use Taylor expansion of $\tan(x)=x+\dfrac {x^3}3+o(x^3)$ and the one of $\dfrac 1{1+u}=1-u+o(u)$
You get $f(x)=\dfrac 9x-\dfrac 9{x+\dfrac {x^3}3+o(x^3)}=\dfrac 9x\left(1-\dfrac 1{1+\dfrac{x^2}3+o(x^2)}\right)=\dfrac 9x\left(1-(1-\dfrac{x^2}3+o(x^2))\right)=3x+o(x)$

Thus $\lim\limits_{x\to 0}f(x)=0$
A: $$\dfrac{\sin x-x\cos x}{x\sin x}=\dfrac{\sin x-x+x(1-\cos x)}{x\sin x}$$
$$=-\dfrac{x-\sin x}{x^3}\cdot\dfrac x{\sin x}\cdot x+\dfrac{1-\cos x}{\sin x}$$
See Are all limits solvable without L'Hôpital Rule or Series Expansion for the first part
and for the second $$\dfrac{1-\cos x}{\sin x}=\dfrac{\sin x}{1+\cos x}$$
