Evaluate the limit $\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{x-\frac{\pi}{4}}$

Question

I am trying to evaluate $$\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{x-\frac{\pi}{4}}$$ without using L'hopital's rule. However, I am not sure what to do. The only thing that came to my mind was to change the tan to sin over cos and get a common denominator but I felt that won't get me anywhere. A hint will be greatly appreciated.

Answer 1

If you substitute $t=x-\pi/4$, then $$ \tan x=\tan(t+\pi/4)=\frac{\tan t+1}{1-\tan t} $$ so your limit is $$ \lim_{t\to0}\frac{1}{t}\left(1-\frac{\tan t+1}{1-\tan t}\right)= \lim_{t\to0}\frac{\tan t}{t}\frac{-2}{1-\tan t} $$ Alternatively, recall that $$ \cos x-\sin x=-\sqrt{2}\sin\left(x-\frac{\pi}{4}\right) $$ and therefore $$ 1-\tan x=-\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\frac{1}{\cos x} $$ Hence the limit can be rewritten as $$ \lim_{x\to\pi/4}\frac{\sin\left(x-\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}\frac{-\sqrt{2}}{\cos x} $$

Answer 2

Hint

This kind of problems are rather simple to address if you know Taylor series. Assuming you do, built at $t=a$, you have

$$\tan(x)=\tan (a)+ \left(\tan ^2(a)+1\right)(x-a)+O\left((x-a)^2\right)$$

Answer 3

We're looking for

$$\lim_{x\to{\pi\over 4}}-{\tan{x}-\tan{\pi\over 4}\over x-{\pi\over 4}}$$

And this is $-\tan'{\pi\over 4}=-2$

Answer 4

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