lim
x-> -infinity
1/(sqaure root of 4/x^2 plus one)
i don't understan why the answer is 1.
as x approaches infinity, the term 4/x^2 becomes a very small number. So, the formula basically becomes:
1/square root(1), which equals 1.
x-> -infinity
1/(sqaure root of 4/x^2 plus one)
i don't understan why the answer is 1.
1/square root(1), which equals 1.