Lazy Limits 2!

Calculus Level 3

$\large \lim_{x\to\pi /4} (\tan x)^{\tan (2x)} = \, ?$

Clarification : $e \approx 2.71828$ denotes the Euler's number .

e^{1}

None of these choices

e^{2}

e^{-2}

1 solution

Chew-Seong Cheong
Jul 8, 2016

$\begin{aligned} \mathscr L & = \lim_{x \to \frac \pi 4} (\tan x)^{\tan (2x)} \\ & = \color{#3D99F6}{\exp} \left[ \ln \left( \lim_{x \to \frac \pi 4} (\tan x)^{\tan (2x)} \right) \right] \quad \quad \small \color{#3D99F6}{\exp (z) = e^z} \\ & = \exp \left[ \lim_{x \to \frac \pi 4} \ln (\tan x)^{\tan (2x)} \right] \\ & = \exp \left[ \lim_{x \to \frac \pi 4} \tan (2x) \ln (\tan x) \right] \\ & = \exp \left[ \lim_{x \to \frac \pi 4} \color{#3D99F6}{\frac {\ln (\tan x)}{\frac 1{\tan (2x)}}} \right] \quad \quad \small \color{#3D99F6}{\text{As this is a 0/0 case, L'Hôpital's rule applies.}} \\ & = \exp \left[ \lim_{x \to \frac \pi 4} \color{#3D99F6}{\frac {\frac {1+\tan^2 x}{\tan x}}{-\frac {2(1+\tan^2 (2x))}{\tan^2 (2x)}}} \right] \quad \quad \small \color{#3D99F6}{\text{Differentiating up and down w.r.t. }x} \\ & = \exp \left[ \lim_{x \to \frac \pi 4} \frac {\frac {1+\tan^2 x}{\tan x}}{-2\left(\frac 1{\tan^2(2x)} + 1 \right)} \right] \\ & = \exp \left[ \frac 2{-2} \right] = e^{-1} \end{aligned}$

Therefore, the answer is: $\boxed{\text{None of the others.}}$

Lazy Limits 2!

1 solution

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