Let $X_1, X_2, X_3,...$ be a sequence of nonnegative random variables such that $\lim\limits_{n\to\infty} E[X_n]=0$.
I want to know if the sequence $Y_n=1-e^{-X_n}$ converges in probability and if so, what is the limit.
My thought is to approach this problem using the Markov inequality:
$$P[X>\epsilon]\leq\frac{E[X]}{\epsilon}$$
Then, by plugging in $Y_n$, I get,
$$P[Y_n>\epsilon]\leq\frac{E[1-e^{-X_n}]}{\epsilon}$$
and since $\lim\limits_{n\to\infty} E[X_n]=0$, then $E[1-e^{-X_n}]$ must also go to $0$ as $n\to\infty$. Therefore, $Y_n$ converges in probability to $0$.
Is this a correct proof?
It is well known that $$0\leq 1-e^{-x}\leq x$$ for all $x\geq 0$ (this fact can be easily proven using Mean Value Theorem for instance). In the particular problem you are considering the inequality translates to $$0\leq Y_n \leq X_n$$ That gives the result $$\lim_{n\to\infty} \mathbb E|Y_n|=0$$ Hence $Y_n$ converges to $0$ in $L^1$ so it converges in probability as well (by Markov's inequality). You indeed said the result; here is a way to prove it.