Convergence in probability: show sequence converges I am having trouble figuring out how to work with convergence in probability questions. I will give a self-study example from Casella and Berger's Statistical Inference Below. 

Let $X_{1}...X_{n}$ be a sequence of random variables that converges
  in probability to a constant a. Assume that $\mathbb{P}(X_i > 0 ) = 1$
  for all i. Verify that the sequences defined by $Y_{i}^{'} = a/X_{i}$
  converge in probability.

I will give the solution to this problem below and then point out where I am having trouble. 
$\mathbb{P}(|a/X_{i} - 1| \leq \epsilon)$ $=$ $\mathbb{P}({{a}\over{1+\epsilon}} \leq X_{i} \leq {{a}\over{1 - \epsilon}})$ = $\mathbb{P}(a-{{a\epsilon}\over{1 + \epsilon}} \leq X_{i} \leq a + {{a\epsilon}\over{1 - \epsilon}})$
( ** In the step above, why is 1 chosen in the expression $(|a/X_{i} - 1|)$? Also, why is $(a - {{a\epsilon}\over{1 + \epsilon}})$ used? I know that since we know $X_{n} \longrightarrow a$ in probability, we're probably trying to rewrite the expression in such a way to use that fact.... but I'm not sure how to just "pull it out of my hat" essentially; ie, in a testing situation. ** ) 
Use $(1 + \epsilon)^{-1} < (1 - \epsilon)^{-1}$ and continue
$\mathbb{P}(|Y_{i} - a^{1/2}| > \epsilon)$ $\geq$ $\mathbb{P}(a-{{a\epsilon}\over{1 + \epsilon}}$ $\leq$ $a + {{a\epsilon}\over{1 - \epsilon}}) = \mathbb{P}(|X_{i} - a|$ $\leq$ $\epsilon{{a}\over{1 + \epsilon}})$ $\longrightarrow 1$ as i $\longrightarrow$  $\infty$ 
( ** In the above, why is the constant being used now $a^{1/2}$ ?? )
Is there some better, more systematic way to approach these problems? I want to be able to encounter any convergence problem and instantly know the steps to employ, rather than try to rewrite the problem in a tricky way.
 A: To sum up, now that you have gone through the steps:


*

*As $(X_i)$ converges in probability to $a$, then $(X_i/a)$ converges in probability to $1$ since$$\mathbb{P}(|X_i/a-1|<\epsilon)=\mathbb{P}(|X_i-a|<\epsilon\times a)=\mathbb{P}(|X_i-a|<\epsilon^\prime),$$meaning that the first term goes to $1$ as $i$ grows to infinity. Hence,$$\mathbb{P}\left(|X_{i}/a - 1| < \frac{\epsilon}{1+\epsilon}\right)=\mathbb{P}(|X_i-a|<\epsilon^{\prime\prime})$$ goes to $1$ as $i$ grows to infinity for every $\epsilon>0$.

*It thus makes sense to check whether or not $(a/X_i)$ converges in probability to $1$. If the sequence converges to $b$, it can only be $b=1$. 

*By definition, $(a/X_i)$ converges in probability to $1$ if $$\mathbb{P}(|a/X_{i} - 1| < \epsilon)$$ goes to $1$ for every $\epsilon>0$.

*The equation$$\mathbb{P}(|a/X_{i} - 1| < \epsilon)=\mathbb{P}\left(a-{{a\epsilon}\over{1 + \epsilon}} <
    X_{i} < a + {{a\epsilon}\over{1 - \epsilon}}\right)$$holds for all $\epsilon$'s.

*Since $$\mathbb{P}\left(a-{{a\epsilon}\over{1 + \epsilon}} < X_{i} < a + {{a\epsilon}\over{1 - \epsilon}}\right)\ge\mathbb{P}\left(a-{{a\epsilon}\over{1 + \epsilon}} < X_{i} < a + {{a\epsilon}\over{1 + \epsilon}}\right)$$and$$\mathbb{P}\left(a-{{a\epsilon}\over{1 + \epsilon}} < X_{i} < a + {{a\epsilon}\over{1 + \epsilon}}\right)=\mathbb{P}\left(\left|X_{i}- a \right|< {{a\epsilon}\over{1 + \epsilon}}\right),$$we are back at the convergence in probability of $X_i$ to $a$: the last term goes to $1$ as $i$ grows to infinity for all $\epsilon$'s and hence the larger term$$\mathbb{P}(|a/X_{i} - 1| < \epsilon)$$also goes to $1$ as $i$ grows to infinity for all $\epsilon$'s.

