Condition on $X$ in $(0,1)$ with $m=E(X)$ for $\lim\limits_{t\to1}(1-t)E\left[\frac{(X-m)^2}{(m+t(X-m))^2}\right]$ to exist and be finite 
Let $X \in (0,1)$ be a continuous random variable.   What would be a sufficient conditions on $X$ such that
  \begin{align}
\lim_{t \to 1}\ (1-t)\, \mathbb E \left[  \frac{(X-\mathbb E[X])^2}{ \left(\mathbb E[X]+t(X-\mathbb E[X]) \right)^2}\right]
\end{align}
  is finite. 

I tried the monotone convergence theorem but it does not seem to apply. I also spent some time thinking about how to apply dominated convergence theorem.  However, I am not sure how to bound $$(1-t) \frac{(X-\mathbb E[X])^2}{ \left(\mathbb E[X]+t(X-\mathbb E[X]) \right)^2}$$
I have ploted the function
\begin{align}
f(t)= (1-t)\frac{(a-b)^2}{(b+t(a-b))^2}
\end{align}
for $a \in (0,1)$ and $b \in (0,1)$ and it seems to be bounded.  However, not sure how to find and exact bound. 
 A: Assume that $X$ has continuous distribution with a bounded p.d.f. $f$. Also assume that
$$ f(0^+) := \lim_{x\downarrow 0}f(x) $$
exists. Under this assumption, it is not hard to prove that

$$ \lim_{t \uparrow 1} \, (1-t) \cdot \mathbb{E} \left[ \frac{(X - \mathbb{E}X)^2}{(\mathbb{E}X + t(X - \mathbb{E}X))^2} \right] = (\mathbb{E}X)  f(0^+). $$

Indeed, let $m = \mathbb{E}X$ for simplicity and write
\begin{align*}
&(1-t) \cdot \mathbb{E} \left[ \frac{(X - \mathbb{E}X)^2}{(\mathbb{E}X + t(X - \mathbb{E}X))^2} \right] \\
&\hspace{1em} = \int_{0}^{1} (1-t)\frac{(x-m)^2}{(m + t(x - m))^2} f(x) \, dx \\
&\hspace{1em} = \int_{0}^{1} (1-t)(x-m)^2 \left( \int_{0}^{\infty} u e^{-(m + t(x - m))u} \, du \right) f(x) \, dx \\
&\hspace{1em} = \int_{0}^{\infty} (1-t)m e^{-(1-t)mu} \left( \frac{1}{m} \int_{0}^{1} u (x-m)^2 e^{-tux} f(x) \, dx \right) \, du.
\end{align*}
By the dominated convergence, it follows that the inner integral converges under the joint limit as $u\uparrow\infty$ and $t \uparrow 1$:
\begin{align*}
I(t,u)
&:= \frac{1}{m} \int_{0}^{1} u (x-m)^2 e^{-tux} f(x) \, dx \\
&\hspace{2em} = \frac{1}{m} \int_{0}^{u} \left( \frac{x}{u}-m\right)^2 e^{-tx} f\left(\frac{x}{u}\right) \, dx \\
&\hspace{4em} \xrightarrow[u\uparrow\infty \text{ and } t\uparrow 1]{} \int_{0}^{\infty} m e^{-x} f(0^+) \, dx
= m f(0^+).
\end{align*}
Now by the dominated convergence again,
\begin{align*}
(1-t) \cdot \mathbb{E} \left[ \frac{(X - \mathbb{E}X)^2}{(\mathbb{E}X + t(X - \mathbb{E}X))^2} \right]
&= \int_{0}^{\infty} (1-t)m e^{-(1-t)mu} I(t, u) \, du \\
&= \int_{0}^{\infty} e^{-u} I\left(t, \frac{u}{(1-t)m}\right) \, du \\
&\xrightarrow[t\uparrow 1]{} \int_{0}^{\infty} e^{-u} m f(0^+) \, du
= mf(0^+).
\end{align*}
