Problem source

Find the limit, as $n\rightarrow\infty,$ of each of the following.
You should explain your reasoning briefly. 
\begin{alignat*}{4}
\mathbf{(i)\ \ } & \dfrac{n}{n+1}, & \qquad & \mathbf{(ii)\ \ } & \dfrac{5n+1}{n^{2}-3n+4}, & \qquad & \mathbf{(iii)\ \ } & \dfrac{\sin n}{n},\\
\\
\mathbf{(iv)\ \ } & \dfrac{\sin(1/n)}{(1/n)}, &  & \mathbf{(v)}\ \  & (\arctan n)^{-1}, &  & \mathbf{(vi)\ \ } & \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}.
\end{alignat*}

Solution source

\begin{questionparts}
\item \begin{align*}
\lim_{n \to \infty} \frac{n}{n+1} &= \lim_{n \to \infty} \left (1 - \frac{1}{n+1} \right ) \\
&\underbrace{=}_{\text{sum of limits}}  \lim_{n \to \infty} 1 - \lim_{n \to \infty}   \frac{1}{n+1}\\
&= 1
\end{align*} 
\item \begin{align*}
\lim_{n \to \infty} \frac{5n+1}{n^2-3n+4}  &= \lim_{n \to \infty} \frac{5/n + 1/n^2}{1-3/n+ 4/n^2} \\
&\underbrace{=}_{\text{ratio of limits}} \frac{\displaystyle \lim_{n \to \infty}(5/n + 1/n^2)}{\displaystyle \lim_{n \to \infty}(1-3/n+ 4/n^2)} \\
&= \frac{0}{1} = 0
\end{align*}

\item \begin{align*}
&& \lvert \frac{\sin n}{n} \rvert &\leq \frac{1}{n} \quad \quad (n \geq 1) \\
\Rightarrow && \lim_{n \to \infty} \lvert \frac{\sin n}{n} \rvert &\leq \lim_{n \to \infty}\frac{1}{n} \\
&&&= 0\\
\Rightarrow &&  \lim_{n \to \infty} \frac{\sin n}{n} &= 0
\end{align*}

\item  First note that $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} \to 1$, then $\frac1n$ is a sequence converging to zero, therefore $\frac{\sin 1/n}{1/n}$ also must tend to $1$.

\item Note that $\lim_{x \to \infty} \tan^{-1} x = \frac{\pi}{2}$ and since $n$ is a sequence tending to infinity we must have $\lim_{n \to \infty} \tan^{-1} n = \frac{\pi}{2}$

\item \begin{align*}
\lim_{n \to \infty} \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-\sqrt{n}} &= \lim_{n \to \infty} \dfrac{\frac{1}{\sqrt{n+1}+\sqrt{n}}}{\frac{2}{\sqrt{n+2}+\sqrt{n}}} \\
&= \frac12 \lim_{n \to \infty} \dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\ 
&= \frac12 \lim_{n \to \infty} \dfrac{\sqrt{1+2/n}+\sqrt{1}}{\sqrt{1+1/n}+\sqrt{1}}\\ 
&= \frac12
\end{align*}
\end{questionparts}