# $\lim_{x\to a}\frac{a}{x-a}\operatorname{Si}_a(x)$

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$$_a(x) = \int_{a}^{x}f(t) \ dt = F(x)-F(a)$$.

• 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. – Izak Jenko Jan 16 at 20:15
• "I know this is incorrect" - wait, why do you think your work is incorrect? – runway44 Jan 16 at 21:00
• Is my assumption correct? It is because the function is continuous in that interval? I just thought that wasn't rigorous. – Babado Jan 16 at 21:31
• en.wikipedia.org/wiki/… – runway44 Jan 17 at 4:55
• Thanks! I have edited my post. – Babado Jan 17 at 10:11