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I have to calculate the following limit:

$\lim _{ x\to0^+}{x\displaystyle\int _{x}^{1}\frac{\cos(t)}{t^2}}dt$

I know that, after using the limit i get an undefined expression and therefore can use the L'Hospital rule to get the solution but i dont quite understand how i get rid of the Integral (with normal integration by part i get the integral sinus expression which i dont want to use)

If I define $F(x)=\int_{x}^{1}\frac{\cos t}{t^2}\,dt$ I get from the fundamental theorem of Calculus that $F'(x)=-\frac{\cos x}{x^2}$, but how am I supposed to use that?

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  • $\begingroup$ You use the fundamental theorem of calculus by looking at the minus of the integral $\endgroup$ – Elad May 28 '17 at 18:10
  • $\begingroup$ You can use the fact that $$\int_0^1 \frac{1-\cos t}{t^2}\,dt$$ converges. So you need to look at $$x\int_x^1 \frac{1}{t^2}\,dt.$$ $\endgroup$ – Daniel Fischer May 28 '17 at 18:20
  • $\begingroup$ could somebody check the edit please I'm nt finding the mistake $\endgroup$ – johnka May 28 '17 at 19:18
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I spot some problems. First, $$ \frac{d}{dx}\int_x^1\frac{\cos t}{t^2}\,dt=-\frac{\cos x}{x^2}\quad \text{note the sign!} $$ Next, l'Hospital, gives you (you check all details) $$ \lim_{x\to 0^+}\frac{\int_x^1\frac{\cos t}{t^2}\,dt}{1/x}=\lim_{x\to 0^+}\frac{\frac{d}{dx}\int_x^1\frac{\cos t}{t^2}\,dt}{\frac{d}{dx}1/x}= \lim_{x\to 0^+}\frac{-\frac{\cos x}{x^2}}{-1/x^2}=\cos 0=1. $$

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$f(t)=\frac{1-\cos t}{t^2}$ is an entire function, hence clearly $$ \lim_{x\to 0^+}x \int_{x}^{1}f(t)\,dt = 0.\tag{1}$$ It follows that: $$ \lim_{x\to 0^+}x\int_{x}^{1}\frac{\cos(t)}{t^2}\,dt = \lim_{x\to 0^+} x\int_{x}^{1}\frac{dt}{t^2} = \lim_{x\to 0^+} x\left(-1+\frac{1}{x}\right)=\color{red}{1}\tag{2} $$ and we don't even need to invoke de l'Hospital theorem.

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