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we were asked to complete this question as an exercise in my lecture.

$$ \lim_{x\to\infty} \left( \frac{1}{x^2} \int_{t=1}^x \frac{t^2}{t + |\sin t|} \, dt \right) $$

(original image at https://i.stack.imgur.com/fvjsE.png)

Just wondering whether you could give some hints as to where to start?

I was thinking that it had something to do with the fundamental theorem of calculus. And then maybe using l'hopitals rule to determine the limit but I'm not too sure as to when to apply them and where. We haven't done anything like this in class.

Any help would be greatly appreciated.

Thanks in advance.

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L\Hopital's Rule gives the limit easily. The answer is $\frac 1 2$.

Details: let $f(x)=\int_1^{x} \frac {t^{2}} {t+|\sin\, t|}dt$. Then $\lim \frac {f(x)} {x^{2}}=\lim (\frac {x^{2}} {x+|\sin\, x|}) /{2x}=\lim \frac x {2(x+|\sin\, x|)} =\lim \frac 1 {2(1+|\sin\, x|/x)}=\frac 1 2$ because $\frac {|\sin\, x|} x \to 0$ as $ x \to \infty$.

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  • $\begingroup$ How did you get to that conclusion? $\endgroup$ – Joshie Bread Apr 26 at 5:15
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    $\begingroup$ @BrianTung Thank you. I stand corrected. $\endgroup$ – Kavi Rama Murthy Apr 28 at 4:53

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