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Find limits of sequences III


We shall use the following useful formula.

$\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, which is a variation of $(a-b)(a+b)=a^2-b^2$.

Solution:

Part a

Using the formula $\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, We have
$$
s_n=\dfrac{(n^2+1)-n^2}{\sqrt{n^2+1}+n}=\frac{1}{\sqrt{n^2+1}+n}.
$$ Now it is clear that $\lim_{n\to 0} s_n=0$.


Part b

Using the formula $\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, We have
\begin{align*}
s_n
=&\ \dfrac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}\\
=&\ \frac{n}{\sqrt{n^2+n}+n}\\
=&\ \frac{1}{\sqrt{1+\frac{1}{n}}+1}.
\end{align*} In the last step, we simultaneously divide numerator and denominantor by $n$. Now it is clear that $\lim_{n\to 0} s_n=\dfrac{1}{2}$.


Part c

Using the formula $\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, We have
\begin{align*}
s_n
=&\ \dfrac{(4n^2+n)-4n^2}{\sqrt{4n^2+n}+2n}\\
=&\ \frac{n}{\sqrt{4n^2+n}+2n}\\
=&\ \frac{1}{\sqrt{4+\frac{1}{n}}+2}.
\end{align*} In the last step, we simultaneously divide numerator and denominantor by $n$. Now it is clear that $\lim_{n\to 0} s_n=\dfrac{1}{4}$.


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