Same limit iff $d(x_n,y_n) = 0 $ If $(M, d)$ is complete, prove that two Cauchy sequences $(x_n)_n$ and $(y_n)_n$ have the 
same limit if and only if $d(x_n,y_n) \rightarrow 0.$
My Method : 
For forward implication , 
$$ d(x_n,a)  < \frac{\varepsilon}{2},$$
$$ d(y_n,a)  < \frac{\varepsilon}{2}.$$
Using the triangle  inequality, we have
$$d(x_n,y_n)  < \varepsilon,$$
and therefore $d(x_n , y_n ) \rightarrow 0$. 

Is this right? 
But, for the reverse implication, I used the triangle inequality and cannot proceed further .

Forward :
Let $x_n$ and $y_n$ be two Cauchy Sequences in $(M,d)$. Also, let $x_n \to L$ and $y_n \to L$ with $L \in M$. This indeed means that :
$$d(x_n,L) \to 0 \quad \text{and} \quad d(y_n,L) \to 0 $$
This can equally be expressed as that $\exists \varepsilon >0$ such that 
$$d(x_n,L) < \varepsilon/2 \quad \text{and} \quad d(y_n,L) < \varepsilon/2$$
as $\varepsilon$ can become arbitrary small.
But $d$ is a metric in the space $M$ and thus the Triangle Inequality holds :
$$d(x_n,y_n)  \leq d(x_n,L) + d(y_n,L) < \varepsilon \implies d(x_n,y_n) \to 0$$
Reverse :
Let $x_n$ and $y_n$ be two Cauchy Sequences in $(M,d)$, such that $d(x_n,y_n) \to 0$. Now, let $x_n \to x$ and $y_n \to y$ with $x,y \in M$. We are interested to see if their limits coincide, aka if $x$ and $y$ can be brought infinitely close. For this, we will work over their distance under the metric $d$, namingly $d(x,y)$. But then, it is :
$$d(x,y) \leq d(x,x_n) + d(x_n,y_n) + d(y_n,y) $$
Now, note that since $x_n \to x$ it is $d(x_n,x) \to 0$ and the same holds for $d(y_n,y)$. We have also let that $d(x_n,y_n) \to 0$. Thus, it is $d(x,y) \to 0$ and that essentialy means that $x \equiv y$.
Reference/Note : A nice post on why the limit exists.
