Does $\lim\limits_{t \to 0^+} \int^1_0 tx^{t−1}f(x)dx=f(0)$?

Question

Assume $f$ is Riemann integrable on $[0,1]$ and continuous at $x = 0$. Show that $\lim\limits_{t \to 0^+} \int^1_0 tx^{t−1}f(x)dx=f(0)$.

Hints are welcome.

Answer 1

Don't know the level of rigor needed in your class, but here is the reason why this relation holds:

Integrate by parts: Let $u=f(x)$ and $dv = t x^{t-1} dx$ $$ \int_0^1 t\,x^{t-1}dx = \int_0^1 u\,dv = (u \,v)|_{x=1} - (u \,v)|_{x=0} - \int_0^1 v\,du \\= (x^t f(x) )|_{x=1} - (x^t f(x))|_{x=0} - \int_0^1 x^t f'(x) \,dx = 1^tf(1)-0^t\,f(0) - \int_0^1 x^t f'(x) \,dx $$ Now we pass to the limit in $t$. $1^t f(1) \rightarrow f(1)$. Assuming $f$ is continuous and finite at zero (which by the way is also needed for integration by parts to be valid) $0^t f(0) \rightarrow 0$. And $$- \int_0^1 x^t f'(x) \,dx \rightarrow - \int_0^1 1 f'(x) \,dx = -(f(1) - f(0)) $$ Add the three non-zero terms to get $f(1) - f(1)+f(0) = f(0)$.

I'm not completely comfortable with this, because the reasoning ought to break down when the function is not Reimann integrable (but might be Lesbegue integrable) and my "proof" doesn't seem to break down there, but the idea should be clear enough.