$f:[0,1]\rightarrow\mathbb{R}\in C[0,1]$ show $\lim_{a\rightarrow 0^+}(\int^1_a(t^{\frac{-1}{2}}f(t))\,dt)$ exists $f:[0,1]\rightarrow\mathbb{R}\in C[0,1]$ (is continuous) 
Show that
$$\lim_{a\rightarrow 0^+}(\int^1_a(t^{\frac{-1}{2}}f(t))\, dt)$$ exists
I've thought about using integration by parts, but the problem of 1/sqrt(0) still happens and I don't think continuity is enough to change that.
 A: Since $f$ is continuous on $[0,1]$ so are the functions $g: [0,1]\to \mathbb{R},\ t \mapsto f(t^2)$ and $G:[0,1] \to \mathbb{R},\ G(x)=\int_x^1g(t)\,dt$.
Setting
$$
u=\sqrt{t},
$$
we have
$$
\int_a^1t^{-1/2}f(t)\,dt=2\int_{\sqrt{a}}^1f(u^2)\,du=2\int_{\sqrt{a}}^1g(u)\,du=2G(\sqrt{a}).
$$
Hence
$$
\lim_{a\to0+}\int_a^1t^{-1/2}f(t)\,dt=2\lim_{a\to0+}G(\sqrt{a})=2G(0).
$$
A: this question belongs to the subject of improper integrals.
More generally, if a function $g(t)$ is continuous on $(0, + \infty)$ and there are numbers $a>0, C>0$ such that
$|g(x)|< C x^{a-1}$ for $0<x<A$, then the limit of
$$\int_{\epsilon}^1 g(t) dt$$ exists.  A change of variables $t=x^a$ works, but one also can perform some 'epsilonics':
$$|\int_{\epsilon_1}^{\epsilon_2} g(t) dt|\leq \int_{\epsilon_1}^{\epsilon_2}|g(t)|\leq \int_{\epsilon_1}^{\epsilon_2} C t^{a-1} dt,$$
which in turn equals $\frac{C}{a} (\epsilon_2^a-\epsilon_1^a)$, for $0<\epsilon_1<\epsilon_2.$  The question is thus settled.
The function $g(t)= log(t)$ also belongs to this class.
