In short: no, you did not do it correctly. The reason is that $\infty\cdot 0$ is one of the indeterminate forms, you cannot conclude it's equal to zero.
The rationale being: it can be anything. For instance, $$\begin{align}
\underbrace{n}_{\to \infty}\cdot \underbrace{\frac{1}{n}}_{\to 0} &\xrightarrow[n\to\infty]{} 1\\
\underbrace{n^2}_{\to \infty}\cdot \underbrace{\frac{1}{n}}_{\to 0} &\xrightarrow[n\to\infty]{} \infty\\
\underbrace{n}_{\to \infty}\cdot \underbrace{\frac{1}{n^2}}_{\to 0} &\xrightarrow[n\to\infty]{} 0\\
\underbrace{n}_{\to \infty}\cdot \underbrace{\frac{\cos n}{n}}_{\to 0} &\xrightarrow[n\to\infty]{} \text{ nothing (no limit)}
\end{align}$$
To solve your indeterminate form, you have to use other techniques to "remove" that issue. Taylor expansions, multiplication by a conjugate, L'Hôpital rule... there are known and systematic methods to do so.
E.g., with Taylor expansions: when $u\to 0$,
$$
\sqrt{1-5u} = 1-\frac{5}{2}u + o(u)
$$
so applying it with $u\stackrel{\rm def}{=}\frac{1}{n}\xrightarrow[n\to\infty]{}0$,
$$\begin{align}
n\left(1- \sqrt{1-\frac{5}{n}}\right)
&= \frac{1}{u}\left(1- \sqrt{1-5u}\right)
= \frac{1}{u}\left(1- 1 + \frac{5}{2}u + o(u)\right)
= \frac{1}{u}\left(\frac{5}{2}u + o(u)\right)\\
&= \frac{5}{2} + o(1) \xrightarrow[n\to\infty]{}\frac{5}{2}.
\end{align}$$
Addendum: why does your calculator do that? The correct answer is indeed $\frac{5}{2}=2.5$. Yet, for "very big $n$", your calculator will return $0$ instead... and that boils down to machine precision. I assume it starts by computing the term $\sqrt{1-\frac{5}{n}}$, then plugs in the result to get $1-\sqrt{1-\frac{5}{n}}$, and finally multiplies what remains by $n$ and outputs the value. But for very large $n$, due to machine precision and rounding errors, the first step will be so close to $1$ that the calculator will just compute $1$ instead of $\sqrt{1-\frac{5}{n}}\approx 1 - \frac{5}{2n}$. So then, $1-1=0$, and no matter what the last step should do, it returns $n\cdot 0=0$.