Limit of $\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $ without l'hopital's

Question

Find limit of

$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $$

I started by defining $\ x = y + \frac{\pi}{4} $

$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} = \lim_{y \to 0} \frac{\tan (y+ \pi/4)-1}{y}= ? $$

This is exactly the same question where the solution suggested also used $\ x = y + \pi/4 $ yet I couldn't understand this equality: $$\ \tan(x) = \tan(y + \pi/4) = \frac{\tan y -1 }{1- \tan y} $$ and then how is that exactly equal to limit of $$\ \lim_{y \to 0}\frac{1}{y} \left( 1-\frac{\tan y +1}{1 - \tan y}\right)$$

Answer 1

As an alternative without derivatives, let $x=\frac{\pi}{4}-y$ then

$$\lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4}=\lim_{y \to 0} \frac{\tan \left(\frac{\pi}{4}-y\right)-1}{-y}=\lim_{y \to 0} \frac{1-\frac{\cos \left(2y\right)}{1+\sin \left(2y\right)}}{y}=\lim_{y \to 0} \frac{1-\cos \left(2y\right)+\sin \left(2y\right)}{y(1+\sin \left(2y\right))}=$$

$$=\lim_{y \to 0} \frac{1}{1+\sin \left(2y\right)}\left(4y\frac{1-\cos \left(2y\right)}{4y^2}+2\frac{\sin \left(2y\right)}{2y}\right)=2$$

Answer 2

$$\frac{\tan\left(y+\dfrac\pi4\right)-1}y=\frac{\dfrac{\tan y+\tan\dfrac \pi4}{1-\tan y\tan \dfrac \pi4}-1}y=2\frac{\tan y}{y}\frac1{1-\tan y}\to 2$$

Answer 3

As $\lim_{x\to \pi/4}(1+\tan x)=2$, you can write

$$\lim_{x\to\pi/4}\frac{\tan x-1}{x-\dfrac\pi4}=2\lim_{x\to\pi/4}\frac{\dfrac{\tan x-1}{1+\tan x}}{x-\dfrac\pi4}=2\lim_{x\to\pi/4}\frac{\tan\left(x-\dfrac\pi4\right)}{x-\dfrac\pi4}.$$

Answer 4

By the definition of derivative with $f(x)= \tan x \implies f'(x)=1/\cos^2 x$

$$\lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4}=f'(\pi/4)=2$$

Answer 5

Below is the incomplete representation of the equality, $$\ \tan(x) = \tan(y + \pi/4) $$ This is the correct representation, $$\lim_{x\to {π\over 4}} \tan(x) =\lim_{y\to 0}\tan(y + \pi/4)$$

The left and right limits are equal and not the functions