what's the limit as n goes to infinity of (2 + 4/n^2)^(1/3)
as n --> ∞ , 4/n^2 ---> 0
so your expression becomes
(2+0)^(1/3)
which is the cuberoot of 2 or appr. 1.2599
thnx
To find the limit as n goes to infinity of the expression (2 + 4/n^2)^(1/3), we can use the concept of limits.
Let's simplify the expression first. We can rewrite it as (2/n^2 + 4/n^2)^(1/3), and simplify inside the brackets: (6/n^2)^(1/3).
Now, let's compute the limit as n approaches infinity for this simplified expression. As n goes to infinity, the value of 1/n^2 becomes smaller and smaller, approaching zero. Therefore, as n approaches infinity, the value of (6/n^2) also approaches zero.
Now, let's find the limit of (6/n^2)^(1/3) as n goes to infinity. Taking the cube root of a number approaching zero will still result in zero. Hence, the limit as n goes to infinity of (6/n^2)^(1/3) is 0.
In conclusion, the limit as n goes to infinity of (2 + 4/n^2)^(1/3) is equal to 0.