Determining the convergence of $ \sqrt{n} \sin(\pi/\sqrt{n}) $ I'm trying to find what this function converges to.
$$ \sqrt{n} \sin(π/\sqrt{n}) $$
I've taken the limit:
$$ \lim_{n\rightarrow\infty} \sqrt{n}\sin(π/\sqrt{n}) $$
It appears as though the function takes the following form as n approaches infinity:
\begin{align*}
\lim_{n\rightarrow\infty} \infty \sin(π/\infty) &= \lim_{n\rightarrow\infty} \infty\cdot \sin(0)\\
&= \infty
\end{align*}
I have been told that this converges to $\pi$, which means I'm doing something wrong here. What am I doing wrong? 
 A: Recall that
$$\lim_{x \to 0} \dfrac{\sin(ax)}{x} = a$$
A: Hint:
Let $t=\frac{\pi}{\sqrt{n}}$, use $$ \lim_{x\rightarrow 0} \frac{\sin x}{x} =1$$
A: If your limit seems to be $\infty \times 0$, then you can't just conclude it's $\infty$ or $0$ or some definite value. The result depends on how strongly each factor approaches its limit.
As others suggest, the main thing is to substitute $x=\frac{1}{\sqrt{n}}$. I'd just like to add how to proceed if you don't know that $\lim_{x\to 0}\frac{\sin ax}{x}=a$.
Limits like this are easily solved by the L'Hôpital's_rule. Let's rewrite
$$
\lim_{n\to\infty}\sqrt{n}\sin(\pi/\sqrt{n})
= \lim_{x\to 0}\frac{\sin(\pi x)}{x} 
$$
Since both the numerator and denominator approach $0$ as $x\to 0$ and we can apply the rule:
\begin{align}
\lim_{x\to 0}\frac{\sin(\pi x)}{x} 
&= \lim_{x\to 0}\frac{\frac{\partial}{\partial x}\sin(\pi x)}{\frac{\partial}{\partial x}x} \\
&= \lim_{x\to 0}\frac{\pi \cos(\pi x)}{1}  \\
&= \pi\cos 0 \\
&= \pi.
\end{align}
Since the limit on the RHS exists, we can conclude that the original limit is also $\pi$.

It'd be also possible to compute the limit using the rule directly, without any "smart" substitution. If we express it as
$$
\lim_{n\to\infty}\sqrt{n}\sin(\pi/\sqrt{n}) =
\lim_{n\to\infty}\frac{\sin(\pi n^{-1/2})}{n^{-1/2}}
$$
we see that the numerator and denominator both approach $0$. Using the rule we compute
\begin{align}
\lim_{n\to\infty}\frac{\frac{\partial}{\partial n}\sin(\pi n^{-1/2})}{\frac{\partial}{\partial n}n^{-1/2}}
&= \lim_{n\to\infty}\frac{-\frac{1}{2}\pi n^{-3/2}\cos(\pi n^{-1/2})}{-\frac{1}{2}n^{-3/2}} \\
&= \lim_{n\to\infty}\pi\cos(\pi n^{-1/2}) \\
&= \pi 
\end{align}
A: Try this question by writing the Taylor series for $\sin(x)$.
$$\sin(x)=\sum_{n=0}^{+\infty}\frac{x^{2n+1}}{(2n+1)!}$$
When you apply this expansion to the question, $\sqrt(n)$ gets cancelled from the numerator and the denominator thus leaving you with one term of the series independent of $n$. Now when you apply the limit, all the other terms tend to zero, as the term containing $n$ is in the denominator. As $n\to\infty$, all the terms tend to zero and the limit converges at $\pi$.
