Answer:
![\displaystyle \lim _(n\to \infty )\sqrt[n]{(n^2+1)/(n+1)} = \boxed1](https://img.qammunity.org/2022/formulas/mathematics/high-school/rkzhligzplfvhbk1s4j6embd167s3cchby.png)
Explanation:
we want to compute the following limit:
![\displaystyle \lim _(n\to \infty )\sqrt[n]{(n^2+1)/(n+1)}](https://img.qammunity.org/2022/formulas/mathematics/high-school/i95zsgzn3jcz61bsm48ovfk64csssmedxq.png)
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]{(n^2)/(n)}](https://img.qammunity.org/2022/formulas/mathematics/high-school/zdca9xol4kk4hx11gn1xk3wyuvb5in9brm.png)
reduce fraction:
![\displaystyle \lim _(n\to \infty )\sqrt[n]{n}](https://img.qammunity.org/2022/formulas/mathematics/high-school/qywqm6bej215sq77ur9rs13eswzr4tik40.png)
by using limit formula, we acquire:
![\displaystyle \lim _(n\to \infty )\sqrt[n]{n} = \boxed1](https://img.qammunity.org/2022/formulas/mathematics/high-school/ozhcp9ocd8rmzsy30q8xgm4lxtxkvyv9l0.png)
and we're done!