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Let $a>0$ and $\operatorname{Si}_a(x)=\int_{a}^{x}\frac{\sin(t)}{t} \, dt$. Compute \begin{equation*} \lim_{x\to a}\frac{a}{x-a}\operatorname{Si}_a(x) \end{equation*} My reasoning was: suppose $F$ is such that $F'(x) = \frac{\sin(x)}{x}$. Then $$a\lim_{x\to a}\frac{\operatorname{Si}_a(x)}{x-a}=a \lim_{x \to 0} \frac{F(x)-F(a)}{x-a} = a F'(a) = a \frac{\sin(a)}{a} = \sin(a)$$

I know this is incorrect, but it gives the correct solution. Does anyone know how to properly solve this?

Edit: Since $f(t)=\frac{sin(t)}{t}$ is continuous in $(0,\infty)$ (in particular is continuous in $(a,x)$), we can use the Fundamental Theorem of Calculus and write Si$_a(x) = \int_{a}^{x}f(t) \ dt = F(x)-F(a)$.

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    $\begingroup$ What seems to be the problem with just taking $F = \operatorname{Si}_a$? Since $\sin(x)/x$ is continuous on $(0, \infty)$, $\operatorname{Si}_a$ is differentiable there. $\endgroup$ – Izak Jenko Jan 16 at 20:15
  • $\begingroup$ "I know this is incorrect" - wait, why do you think your work is incorrect? $\endgroup$ – runway44 Jan 16 at 21:00
  • $\begingroup$ Is my assumption correct? It is because the function is continuous in that interval? I just thought that wasn't rigorous. $\endgroup$ – Babado Jan 16 at 21:31
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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – runway44 Jan 17 at 4:55
  • $\begingroup$ Thanks! I have edited my post. $\endgroup$ – Babado Jan 17 at 10:11

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