Finding the limit of trigonometric function

Question

$$\lim_{x\to\frac{\pi}{4}}\tan(2x)\cdot\tan\left(\frac{\pi}{4}-x\right)$$ How do I find the limit of this function without L'hospital rule.

Answer 1

Hint:

Using the formula $\displaystyle \tan(a + b) = \frac{ \tan a + \tan b}{1 - \tan a . \tan b}$,

$$\tan 2x . \tan (\pi/4 - x) = \frac{2\tan x}{1 - \tan^2 x}.\frac{\tan\pi/4 + \tan(-x)}{1 - \tan\pi/4.\tan(-x)}$$

Simplify that expression and then try and take the limit

Answer 2

Let $t=x-\frac\pi4$ then we write the given limit on this form ($\tan(2t+\pi/2)=-\cot(2t)$):

$$\lim_{t\to0}\frac{\tan t}{\tan 2t}$$ Now the result is clearly $\frac12$ using that

$$\lim_{t\to0}\frac{\sin t}{t}=1$$

Answer 3

simplifying your given term into $2\,{\frac {\sin \left( x \right) \cos \left( x \right) \left( \cos \left( x \right) -\sin \left( x \right) \right) }{ \left( 2\, \left( \cos \left( x \right) \right) ^{2}-1 \right) \left( \cos \left( x \right) +\sin \left( x \right) \right) }} $ and the searched limit is $\frac{1}{2}$

Answer 4

Another approach:

$$\tan2x\;\tan\left(\frac\pi4-x\right)=\frac{\sin2x}{\cos2x}\frac{\sin\left(\frac\pi4-x\right)}{\cos\left(\frac\pi4-x\right)}=\frac{2\sin x\cos x}{\cos^2x-\sin^2x}\frac{\frac1{\sqrt2}\left(\cos x-\sin x\right)}{\frac1{\sqrt2}\left(\cos x+\sin x\right)}=$$

$$=\frac{2\sin x\cos x}{(\cos x+\sin x)^2}\xrightarrow[x\to\frac\pi4]{}0\frac{2\frac1{\sqrt2}\frac1{\sqrt2}}{\left(\frac2{\sqrt2}\right)^2}=\frac12$$