How does one solve this limit $ \lim\limits_{x\to0}\frac1x\int_{0}^{x}[1-\tan{2t}]^{1/t}dt=e^{-2} $?

Question

$$ \lim\limits_{x\to0}\dfrac{1}{x}\int_{0}^{x}[1-\tan{2t}]^{1/t}dt=e^{-2} $$

Notice that

$$ f(x)=\int_0^x[1-\tan{2t}]^{1/t}dt \to 0,\ \text{when}\ x \to 0, $$

and obviously

$$ g(x)=x \to 0,\ \text{when}\ x \to 0 $$

so we can apply L'Hopitals Rule. By the Fundamental Theorem of Calculus, we have

$$ f'(x)=\dfrac{d}{dx}\int_0^x[1-\tan{2t}]^{1/t}dt=[1-\tan{2x}]^{1/x} $$ $$ g'(x)=1 $$

So we obtain

$$ \lim\limits_{x\to0}[1-\tan{2x}]^{1/x}. $$

But what can I do now?

Answer 1

$\lim_{x \to 0} \frac {\ln (1-\tan (2x))} x=\lim_{x \to 0} \frac {-2\sec^{2}(2x)} {1-\tan (2x)}=-2$ by L'Hopital's Rule. Hence $\lim_{x \to 0}(1-\tan (2x))^{1/x}=e^{-2}$.

Answer 2

A hint to continue: Note that $(1-\tan 2x)^{1/x} = e^{\frac{\ln(1-\tan 2x)}{x}}$.