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I need some help with a question. I have to calculate

$$\lim_{x \to 3}\frac{x^2}{x - 3}\int_3^x \frac{\sin t}{t}dt.$$

If I'm not wrong, we can write

$$\sin(x) = \sum_{n = 0}^{\infty}\frac{x^{2n+1}}{(2n+1)!} \Longrightarrow \frac{\sin(x)}{x}=\sum_{n = 0}^{\infty}\frac{x^{2n}}{(2n+1)!},$$

then

$$\int_3^x \frac{\sin (t)}{t}dt = \sum_{n = 0}^{\infty}\int_3^x \frac{t^{2n}}{(2n+1)!}dt = \sum_{n = 0}^{\infty}\frac{t^{2n+1}}{(2n+1)(2n+1)!}\Bigg|_3^x.$$

But I don't know how exactly to use it. I appreciate any help!

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  • $\begingroup$ Are you allowed to use L'Hospital's rule? $\endgroup$
    – Koro
    Dec 2 '21 at 4:52
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    $\begingroup$ @Koro yes!${}{}$ $\endgroup$
    – Greg
    Dec 2 '21 at 4:53
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    $\begingroup$ Your series for $\sin(x)$ is wrong. You miss the factor $(-1)^n$ $\endgroup$ Dec 2 '21 at 5:00
  • $\begingroup$ @ClaudeLeibovici oh, I didn't notice. Thanks! $\endgroup$
    – Greg
    Dec 2 '21 at 5:19
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You can use L'Hospital's rule and the fundamental theorem: $$ \lim\limits_{x\to 3}(...)=\lim\limits_{x\to 3}\frac{2x\int_3^x\frac{\sin t}{t}\,dt+x^2\frac{\sin x}{x}}{1}=3\sin 3. $$

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By L'Hospital's rule and the fundamental theorem of calculus:

$$\lim_{x\to 3} \frac{x^2}{x-3} \int_3^x \frac{\sin t}{t} dt = 9\lim_{x\to 3}\frac{\int_3^x \frac{\sin t}{t} dt}{x-3} = 9\lim_{x\to 3} \frac{\sin x}{x} = 3\sin 3$$

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Using L’Hospital Rule and the Fundamental Theorem of Calculus, we have

$$ \begin{aligned} & \lim _{x \rightarrow 3} \frac{x^{2}}{x-3} \int_{3}^{x} \frac{\sin t}{t} d t \quad\left(\frac{0}{0}\right) \\ =& \lim _{x \rightarrow 3} \frac{x^{2} \frac{\sin x}{x}+2 x \int_{3}^{x} \frac{\sin t}{t} d t}{1} \\ =& \frac{3^{2} \sin 3}{3}+0 \\ =& \frac{\sin 3}{3} \end{aligned} $$

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    $\begingroup$ You have the cancellation backwards between your last two lines. $\endgroup$ Dec 2 '21 at 13:09

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