retagged by
397 views

2 Answers

Best answer
3 votes
3 votes
$$\lim_{x \to \frac{\pi}{4}} (\tan x)^{\tan(2x)} = \lim_{x \to \pi/4} e^{\tan 2x \ln(\tan x)}$$

Now,

$\lim_{x \to\frac{\pi}{4}} \tan (2x)\ln(\tan x) \\= \lim_{x \to\frac{\pi}{4}} \frac{\sin 2x \ln(\tan x)}{\cos 2x}  \left(\frac {0}{0} \text{ form }\right)\\=  \lim_{x \to\frac{\pi}{4}} \frac { \frac{ \sin 2x \sec^2 x}{\tan x} + 2 \cos 2x . \ln \tan x }{-2 \sin 2x} (\text{L'Hospital's rule})\\= \frac{2}{-2}= -1$

So,

$$\lim_{u \to -1} e^u = \frac{1}{e}, \\\text{where } u = \tan (2x) \ln(\tan x)$$
selected by

Related questions

1 votes
1 votes
3 answers
1
Salman asked Aug 1, 2015
889 views
$$\lim_{x \to 0} \frac{\cos(x)-\log(1+x)-1+x}{\sin^2x} = ? $$Please explain the steps also
1 votes
1 votes
1 answer
2
saket nandan asked Jul 10, 2015
401 views
$\lim_{x\rightarrow 0}\frac{ \sin x^{\circ}}{x}$
0 votes
0 votes
1 answer
3
saket nandan asked Jul 10, 2015
384 views
$\lim_{x\rightarrow 0} \left( \frac{1}{x^{2}} -\frac{1}{\sin^{2}x} \right)$
0 votes
0 votes
2 answers
4
saket nandan asked Jul 10, 2015
754 views
$\lim_{\theta \rightarrow 0} \frac{ 1-\cos \theta }{ \theta \sin\theta }$