Let $k$ be some constant. If $$\dfrac{k-n}{\sqrt{n}} \to x \;\;\text{as} \;\;n \to \infty$$ does it follow that $$k-n \to x \sqrt{n} \;\;\text{as} \;\; n \to \infty?$$ I looked at Calculus limit laws, but I'm not finding any insight of whether this type of manipulation is allowed or not. Any feedback would be appreciated! For context, this is from a question from Durrett's Probability Text, Problem 3.1.2. Here's a snippet of the problem: "Use Stirling's formula to show that if $(k-n)/\sqrt{n} \to x$, then $\sqrt{2\pi n}P(S_n =k) \to exp(-x^2/2)$ where $P(S_n=k)=e^{-n}n^k/k!$"
EDIT: I am asking this question because I see the $x\sqrt{n}$ being used in the solution and am having a hard time justifying Durrett's usage of it.
We have \begin{align*} &\sqrt {2\pi n} P(S_n = k) = \sqrt {2\pi n} e^{ - n} \frac{{n^k }}{{k!}} \sim e^{k - n} \left( {\frac{n}{k}} \right)^{k + 1/2} = e^{k - n} \left( {1 + \frac{{n - k}}{k}} \right)^{k + 1/2} \\ & = \exp \left[ {k - n + \left( {k + \frac{1}{2}} \right)\log \left( {1 + \frac{{n - k}}{k}} \right)} \right] \\ & = \exp \left[ {k - n + \left( {k + \frac{1}{2}} \right)\left( {\frac{{n - k}}{k} - \frac{{(n - k)^2 }}{{2k^2 }} + \mathcal{O}\!\left( {\frac{{(n - k)^3 }}{{k^3 }}} \right)} \right)} \right] \\ & = \exp \left[ { - \frac{{(n - k)^2 }}{{2k}} + \mathcal{O}\!\left( {\frac{{(n - k)^3 }}{{k^2 }}} \right) + \frac{{n - k}}{{2k}} - \frac{{(n - k)^2 }}{{4k^2 }} + \mathcal{O}\!\left( {\frac{{(n - k)^3 }}{{k^3 }}} \right)} \right]. \end{align*} By assumption, $k - n = x\sqrt n + o(\sqrt n )$, giving $$ \sqrt {2\pi n} P(S_n = k) = \exp \left[ { - \frac{{x^2 n}}{{2k}} + o\!\left( {\frac{n}{k}} \right) + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right]. $$ Again by assumption, $k = n + \mathcal{O}(\sqrt n )$, giving $$ \sqrt {2\pi n} P(S_n = k) = \exp \left[ { - \frac{{x^2 }}{2} + o(1)} \right]. $$