Help me with this Limit

Question

I don't have idea how to make this limit, i read it in a math contest. I think that is a limit that could be attacked by method of Riemann's sums. $$\lim_{x\to 0} \int _0 ^ {x} (1- \tan (2t) ) ^ {\frac{1}{t}}\ dt$$

Can you help me?

Answer 1

Note that $$|x|<\pi/8\implies 0\leq\left|\int_0^x (1-\tan(2t))^{1/t}dt\right|\leq \int_{-|x|}^{|x|} |1-\tan(2t)|^{1/t}\leq\int_{-|x|}^{|x|} 1dt=2|x|$$ and since $2|x|\to 0$, the conclusion follows by the Squeeze theorem.

Answer 2

First lets observe that

$$\lim_{t \to 0} (1- \tan (2t) ) ^ {\frac{1}{t}}=e^{-2}$$

Thus the function $f(t)=(1- \tan (2t) ) ^ {\frac{1}{t}}$can be extended continuously to $0$ by setting $f(t)=e^{-2}$.

Then by FTC

$$\lim_{x \to 0}\dfrac{\int _0 ^ {x} (1- \tan (2t) ) ^ {\frac{1}{t}}\ dt}{x}= f(0)=e^{-2}$$

which implies $$\lim\limits_{x \to 0}\int _0 ^ {x} (1- \tan (2t) ) ^ {\frac{1}{t}}\ dt =0$$