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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?

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I don't believe this function has an elementary antiderivative, so that doesn't look good. Why is this not simply 0? –  Argon May 9 '12 at 22:10
    
yes, same i thought. –  d555 May 9 '12 at 22:13
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I think the integrand goes to $e^{-2}$ as $t \to 0^+$, so the limit is $0$ –  Stefan Smith May 9 '12 at 22:15
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Doesn't the integrand approach $e^{-2}$ continuously as $t\to 0$? If this is the case, this limit should really just be $0$. –  froggie May 9 '12 at 22:15
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Just note for $|x|$ small, the integrand is positive and bounded above by 1; thus the integral $I$ satisfies $0\le I\le |x|$, when $|x|$ is small. –  David Mitra May 9 '12 at 22:15
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2 Answers

up vote 4 down vote accepted

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.

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@DavidMitra Keen observation. –  Pedro Tamaroff May 9 '12 at 22:28
    
@DavidMitra Thanks for the catch. Dividing by two turned into multiplying in my head. –  Alex Becker May 9 '12 at 22:28
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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 $$\lim\limits_{x \to 0}\int _0 ^ {x} (1- \tan (2t) ) ^ {\frac{1}{t}}\ dt =0$$

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