Compute $\lim\limits_{n \to \infty} \dfrac{5 + (-1)^n}{ \sqrt{n} +7}$

Question

Compute the limit : $\lim\limits_{n \to \infty} \dfrac{5 + (-1)^n}{ \sqrt{n} +7}$

My work:

Since $(-1)^n$ can be either -1 or +1 so, with Sandwich Law the numerator can be within 4 and 6 i.e $$ 4 \leqslant 5 + (-1)^n \leqslant 6$$. The denominator grows accordingly with $\sqrt{n}$. Using this info how can I proceed further ?

Answer 1

Divide the inequality $4 \leqslant 5 + (-1)^n \leqslant 6$ by $\sqrt{n}+7$ to get that $$ \frac{4}{\sqrt{n}+7}\leqslant \frac{5+(-1)^n}{\sqrt{n}+7}\leqslant\frac{6}{\sqrt{n}+7} \, . $$ Since $$ \lim_{n \to \infty}\frac{4}{\sqrt{n}+7}=\lim_{n \to \infty}\frac{6}{\sqrt{n}+7}=0 \tag{*}\label{*} \, , $$ by the sandwich theorem we have $$ \lim_{n \to \infty}\frac{5+(-1)^n}{\sqrt{n}+7}=0 \, . $$ For a proof of $\eqref{*}$, we can show that $$ \lim_{n \to \infty}\frac{c}{\sqrt{n}+7}=0 $$ for any constant $c$ using the formal definition of the limit. For every $\varepsilon>0$, if $\displaystyle{n>\max\left(100,\frac{|c|}{\varepsilon}\right)}$, then $$ \left|\frac{c}{\sqrt{n}+7}\right|=\frac{|c|}{\sqrt{n}+7}<\frac{|c|}{n}<\varepsilon \, . $$

Answer 2

We have $$ \lim\limits_{n \to \infty} \dfrac{5}{ \sqrt{n} +7} =0 $$ and $$ \left |\dfrac{(-1)^n}{ \sqrt{n} +7} \right | \leqslant \dfrac{1}{ \sqrt{n} +7} \xrightarrow[n \to \infty]{} 0 $$ so the limit is $0$.

Answer 3

My approach. Being

$$\lim_{n \to \infty} \frac{5 + (-1)^n}{\sqrt{n}+7}=\color{magenta}{\lim_{n \to \infty} \frac{5}{\sqrt{n} +7}}+\color{blue}{\lim_{n \to \infty} \frac{(-1)^n}{\sqrt{n} +7}}$$ The sequence $(-1)^n$ is bounded. The product of a finite and an infinitesimal sequence gives a null limit. Hence the blue limit is $0$. The sequence is easy to solve because $$\color{magenta}{\lim_{n \to \infty} \frac{5}{\sqrt{n} +7}}\equiv 0$$ Definitively:

$$\lim_{n \to \infty} \frac{5 + (-1)^n}{\sqrt{n}+7}=0$$