# Solve the following (limit; L'Hospital's Rule)?

## $\setminus {\lim}_{x \setminus \rightarrow {0}^{+}} {\left(\setminus \tan \left(2 x\right)\right)}^{x}$ What I've tried $y = {\left(\setminus \tan \left(2 x\right)\right)}^{x}$ $\setminus \Rightarrow \setminus \ln \left(y\right) = \setminus {\lim}_{x \setminus \rightarrow {0}^{+}} \setminus \ln \left({\left(\setminus \tan \left(2 x\right)\right)}^{x}\right) \ldots$ got stuck afterwards.

Nov 26, 2016

$1$

#### Explanation:

${\left(\tan \left(2 x\right)\right)}^{x} = {\left(\sin \left(2 x\right)\right)}^{x} / {\left(\cos \left(2 x\right)\right)}^{x}$

sin(2x)= 2x-(2x)^3/(3!)+(2x)^5/(5!)+ cdots

is an alternate series and for $- \frac{\pi}{2} \le x \le \frac{\pi}{2}$ verifies

2x-(2x)^3/(3!) le sin(2x) le 2x

then

lim_(x->0)(2x-(2x)^3/(3!))^x le lim_(x->0)(sin(2x))^x le lim_(x->0)( 2x)^x

now using the binomial expansion

x^x = (1+(x-1))^x = 1 + x(x-1)+x(x-1)(x-1)^2/(2!)+x(x-1)(x-2)(x-1)^3/(3!)+cdots = 1+xp(x) where $p \left(x\right)$ represents a suitable polynomial. So

${\lim}_{x \to 0} {x}^{x} = {\lim}_{x \to 0} 1 + x p \left(x\right) = 1$

Of course

${\lim}_{x \to 0} {\left(2 x\right)}^{x} = {\left({\lim}_{y \to 0} {y}^{y}\right)}^{\frac{1}{2}} = {1}^{\frac{1}{2}} = 1$

Finally

${\lim}_{x \to 0} {\left(\tan \left(2 x\right)\right)}^{x} = \frac{{\lim}_{x \to 0} {\left(\sin \left(2 x\right)\right)}^{x}}{{\lim}_{x \to 0} {\left(\cos \left(2 x\right)\right)}^{x}} = \frac{1}{1} ^ 0 = 1$

Nov 26, 2016

1

#### Explanation:

Let y = (tan 2x)^x, then

$\lim x \to {0}_{\pm} \ln y$

$= \lim x \to {0}_{\pm} x \ln \tan 2 x$

$= \lim x \to {0}_{\pm} \frac{\ln \tan 2 x}{\frac{1}{x}}$

$= \lim x \to {0}_{\pm} \frac{\left(\ln \tan 2 x\right) '}{\left(\frac{1}{x}\right) '}$

$= \lim x \to {0}_{\pm} - 2 {x}^{2} {\sec}^{2} \frac{2 x}{\tan} \left(2 x\right)$

=lim x to 0_+-((2x)/(sin 2x))((x))/((cos2x)

$= \left(1\right) \frac{0}{1} = 0$. so,

$y \to {e}^{0} = 1$.