Finding $\lim_{x\to\pi/4}\frac{\tan(x)-1}{x-\pi/4}$. find the limit $$\lim_{x\to\pi/4}\frac{\tan(x)-1}{x-\pi/4}$$
direct substitution results in $0/0$
and it seems that there's no way to factor it
When I look at its graph
It's clear that it has the limit 2 as $x$ approaches $\pi/4$
the question is How to factor this function ??
!
 A: Without applying L'Hospital's Rule,
$$\eqalign{
  & \tan (x - {\pi  \over 4}) = {{\tan x - 1} \over {1 + \tan x}} \longrightarrow \tan x - 1 = (1 + \tan x)\tan (x - {\pi  \over 4})  \cr 
  & \mathop {\lim }\limits_{x \to {\pi  \over 4}} {{\tan x - 1} \over {x - {\pi  \over 4}}}  \cr 
  &  = \mathop {\lim }\limits_{x \to {\pi  \over 4}} {{(1 + \tan x)\tan (x - {\pi  \over 4})} \over {x - {\pi  \over 4}}}  \cr 
  &  = \mathop {\lim }\limits_{x \to {\pi  \over 4}} {{(1 + \tan x)(x - {\pi  \over 4})} \over {x - {\pi  \over 4}}}  \cr 
  &  = \mathop {\lim }\limits_{x \to {\pi  \over 4}} (1 + \tan x)  \cr 
  &  = 2 \cr} $$
A: Use L'Hopital's rule.
$\lim_{x\rightarrow \frac{\pi}{4}} \frac{\tan{x}-1}{x-\frac{\pi}{4}}=\lim_{x\rightarrow \frac{\pi}{4}}\frac{1+\tan^2(x)}{1}=2$ 
A: By using Taylor's expansion we have $\tan(x)\simeq 1+2(x-\frac{\pi}{4})$, around $x=\frac{\pi}{4}$, hence we can say:
$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\tan(x)-1}{x-\frac{\pi}{4}}=\lim_{x\rightarrow \frac{\pi}{4}}\frac{1+2(x-\frac{\pi}{4})-1}{x-\frac{\pi}{4}}=2$
A: Because $\tan(\frac{\pi}{4})=1$ this can be thinked as difference quotient. $\lim_{x\rightarrow\frac{\pi}{4}}\frac{\tan\left(x\right)-1}{x-\frac{\pi}{4}}$ is just like $\lim_{x\rightarrow x_0}\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}$ where $f(x)$ is $\tan(x)$ and $x_0$ is $\frac{\pi}{4}$. Now, $\lim_{x\rightarrow\frac{\pi}{4}}\frac{\tan\left(x\right)-1}{x-\frac{\pi}{4}}=f'(x)$. $f'(x)=D(\tan x)=1+\tan^2x$. And then
$f'(\frac{\pi}{4})=1+\tan^2(\frac{\pi}{4})=1+1^2=2$
A: With a shift of the variable ($t\to0$),
$$\frac{\tan(t+\frac\pi4)-1}{t}=\frac{\dfrac{1+\tan t}{1-\tan t}-1}t=\frac{2\tan t}{t(1-\tan t)}\to2.$$
A: It is not necessary to factor! This limit is not so coincidentally the derivative of $f(x)=\tan(x)$ at the point $x=\pi/4$. By a substitution it can be shown $$ f'(a) = \lim_{x \to a }\frac{f(x) - f(a) }{x-a} $$
Specifically $$ \tan'(\pi/4) =\lim_{x\to\pi/4}\frac{\tan(x)-\tan(\pi/4)}{x-\pi/4}$$ Then using the fact that $ \tan (x) ' =\sec^2( x)  $ we immediately get the limit is $$ \sec\left(\pi/4\right)^2 = \left(\frac {2}{\sqrt 2 } \right)^2 = 2 $$
