find $f'(0)$, while $f(x)=\int_{0}^{x} \sin(t)/t \,dt$

Question

I have an integral: $f(x)=\int_{0}^{x} \sin(t)/t\, dt$
And I have to found $f'(0)$

So, I know that I can look at the integrand as at func $ g(x)=\sin(x)/x$ for $x \neq 0 $ $1$ for $x=0$
And the result is $1$

Is it correct to write it like this: $f'(0)=\lim_{x\to 0} (\int_{0}^{x} \sin(t)/t\, dt - \int_{0}^{0} \sin(t)/t \,dt)/(x-0)$ ,
and from L'Hospital rule:
$\lim_{x\to 0} \sin(x)/x =1$

My Questions: I'm not sure if I can use $\int_{0}^{0} \sin(t)/t\, dt=0$ , because the integrand isn't defined at $x=0$

Answer 1

Since $$\left(\int\limits_0^x\frac{\sin{t}}{t}dt\right)'=\frac{\sin{x}}{x},$$ it's just $\lim\limits_{x\rightarrow0}\frac{\sin{x}}{x}=1$.

Answer 2

The value of the integral doesn't change if you change the integrand at a single point, so you can assume the integrand is $$f(t)=\cases{\sin t/t ,&$t\neq0$\\1,&$t=0$}$$ and apply the fundamental theorem of calculus.

Answer 3

If you apply the fundamental theorem of calculus to $f(x)$ you will get that $f'(x)=\frac{\sin x}{x}$ and from there you take the limit in the answer by Michael.

Answer 4

You do not explicitly mention where $f(x)$ is defined, but let us consider it as a function $f : \mathbb R \to \mathbb R$.

The first question is whether $f(x)$ is well-defined, that is, whether $\int_0^x \frac{\sin t}{t} dt$ exists. You correctly say it is the same as $$\int_0^x g(t)dt$$ with $g(x) = \begin{cases} \frac{\sin x}{x} & x \ne 0 \\ 1 & x = 0 \end{cases} \quad.$ But $g$ is continuous. This is obviuos in all $x \ne 0$. In $x = 0$ it follows from $\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\sin x -\sin 0}{x - 0} = \sin'(0) = cos 0 = 1$.

Hence $f(x)$ is well-defined and the fundamental theorem of calculus implies that $f$ is differentiable with $f' = g$. Hence $f'(0) = g(0) = 1$.