I'm having trouble finding
$$
\lim_{x\to0}\biggl(\frac{1}{x}-\frac{1}{\tan x}\biggr)
$$
As $x$ approaches zero, the function is in indeterminate form, so I used L'Hôpital's rule and took the derivative:
$$-\frac{1}{x^2}-\frac{\sec^2x}{\tan^2x}
$$
This is still in indeterminate form and I still can't take the limit as x approaches zero, so I took the derivative again and still got it in indeterminate form. Is there something that I'm doing wrong either algebraically or with respect to differentiation?
Best Answer
As mentioned by @onamoonlessnight in the comments, L'Hopitals works for a ratio of functions both tending to $0$ or $\pm\infty$. First combine:
$$\dfrac{1}{x}-\dfrac{1}{\tan x} = \dfrac{\tan x - x}{x\tan x} \to \ \dfrac{0}{0}$$
Now use L'Hopital's Rule.
$$\lim_{x\to 0} \dfrac{\tan x - x}{x\tan x} =_{L'H} \lim_{x\to 0} \dfrac{\overbrace{\sec^2x-1}^{\tan^2 x}}{x\sec^2x + \tan x} = \lim_{x\to 0} \dfrac{\sin x}{\dfrac{x}{\sin x} + \cos x} $$