Let $X=(0,1]$ and

$$d\_1(x,y)=|x-y|, \ d\_2(x,y)=|\frac{1}{x}-\frac{1}{y}|$$

be two metric on $X$. Show that $(X,d\_1), (X,d\_2)$ have the same convergent sequences, but $(X,d\_2)$ is complete while $(X,d\_1)$ is not.

Attempt:
Let $x\_n$ be convergent in $(X,d\_1)$
$\Longrightarrow |x\_n-x|\rightarrow 0,n\rightarrow\infty $ 
$\Longrightarrow |\frac{1}{x\_n}-\frac{1}{x}|=\frac{|x-x\_n|}{x\_n\*x}=d\_2(x\_n,x)\rightarrow 0, n\rightarrow\infty$ [is this enough to prove they have the same convergent sequences?]

To show that $(X,d\_1)$ is not complete, consider sequence $\frac{1}{n}$, which is Cauchy, but converges to $0\notin X, n\rightarrow\infty$

[This is the primary source of my troubles:]
I am trying to use an isometry: $\phi:(0,1]\to[1,\infty)$ defined by $\phi(t)=\frac{1}{t}$ to show that $(X,d\_2)$ is complete.

I think I can show that $\phi(x\_n)$ converges to y in $[1,\infty)$ and therefore, $x\_n$ converges to $\frac{1}{y} \in X$. but, I am having trouble in the execution.