Question

${lim}_{n\to \infty }\frac{{(\sqrt{n^2+1}+n)}^2}{\sqrt[3]{3n^6+1}}=$

• A. $2$
• B. $1/2$
• C. $1/3$
• D. $1/4$

Answer 1

The correct option is A $2$

${lim}_{n\to \infty }\frac{{(\sqrt{n^2+1}+n)}^2}{\sqrt[3]{3n^6+1}}$

Taking $n^2$ common from both numerator and denominator

$={lim}_{n\to \infty }\frac{n^2(\sqrt{1+\frac{1}{n^2}}+1)}{n^2\left(\sqrt[3]{31+\frac{1}{n^6}}\right)}$

$={lim}_{n\to \infty }\frac{(\sqrt{1+\frac{1}{n^2}}+1)}{\sqrt[3]{31+\frac{1}{n^6}}}$

$=\frac{1+1}{1}$

$=2$

Hence, the answer is $2.$