Finding $\lim_{x \rightarrow \frac{1}{4} \pi } \frac{\tan x-\cot x}{x-\frac{1}{4} \pi }$.

Question

How do I get the value of

$$\lim_{x \rightarrow \frac{1}{4} \pi } \frac{\tan x-\cot x}{x-\frac{1}{4} \pi }?$$

I need the steps without using L'hospital.

Answer 1

$$...=\underbrace{\frac{1}{\tan x}}_{\to 1}\underbrace{\frac{\tan^2x-1}{(x-\frac{\pi}{4})}}_{ \to \frac{d}{dx}\tan^2(x)\big|_{x=\pi/4}}$$

Answer 2

idm's answer is good, but L´Hôpital's rule is often more directly an instance of a limit actually being the limit of a difference quotient defining a derivative, or, in this case, a pair of difference quotients defining a pair of derivatives:

$$\frac{\tan x-\cot x}{x-\frac{1}{4} \pi } = \frac{\tan x - 1}{x-\frac{1}{4} \pi } - \frac{\cot x -1}{x-\frac{1}{4} \pi }.$$

then we have

$$\lim_{x \rightarrow \frac{1}{4} \pi } \frac{\tan x-\cot x}{x-\frac{1}{4} \pi } = \lim_{x \rightarrow \frac{1}{4} \pi} \frac{\tan x - 1}{x-\frac{1}{4} \pi } - \frac{\cot x -1}{x-\frac{1}{4} \pi }$$ $$=\tan'\left(\frac{1}{4}\pi\right) - \cot'\left(\frac{1}{4}\pi\right)$$

Answer 3

Firs note that $ \displaystyle \tan x - \cot x = - 2 \frac{\cos 2x}{\sin 2x}$. Then consider a change of variable $\displaystyle u=x-\frac{\pi}{4}$ to obtain

$$ \lim_{x \rightarrow \frac{1}{4} \pi } \frac{\tan x-\cot x}{x-\frac{1}{4} \pi }= \lim_{u \rightarrow 0 } [\frac{\sin 2u}{u}\times \frac{2}{\cos2u}]=\lim_{u \rightarrow 0 } [\frac{\sin 2u}{u}] \times \lim_{u \rightarrow 0 } [ \frac{2}{\cos2u}].$$ Now note that $ \displaystyle \lim_{u \rightarrow 0 } \frac{\sin 2u}{u} = 2 \lim_{u \rightarrow 0 } \frac{\sin 2u}{2 u} = 2 \lim_{t \rightarrow 0 } \frac{\sin t}{t} =2$, which has received some good bit of attention here.

The limit will hence be $\displaystyle 2 \times 2 =4$.