Question

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

Please show your work!
Thank you! :)

Answer 1

Answer:

$\displaystyle \lim _{n\to \infty }\sqrt[n]{\frac{n^2+1}{n+1}} = \boxed1$

Step-by-step explanation:

we want to compute the following limit:

$\displaystyle \lim _{n\to \infty }\sqrt[n]{\frac{n^2+1}{n+1}}$

well, remember that, for limits to infinity, terms less than the highest degree of the numerator or denominator can be disregarded, hence we can drop 1 which yields:

$\displaystyle \lim _{n\to \infty }\sqrt[n]{\frac{n^2}{n}}$

reduce fraction:

$\displaystyle \lim _{n\to \infty }\sqrt[n]{n}$

by using limit formula, we acquire:

$\displaystyle \lim _{n\to \infty }\sqrt[n]{n} = \boxed1$

and we're done!