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Limit involving square roots?
How do you do this limit? Thanks in advance!
lim x -> infinity of sqrt(x^2 + x) - x
2 Answers
- 5 years agoFavorite Answer
Rationalize the numerator
(sqrt(x^2 + x) - x) / 1 =>
(sqrt(x^2 + x) - x) * (sqrt(x^2 + x) + x) / (1 * (sqrt(x^2 + x) + x)) =>
(x^2 + x - x^2) / (x + sqrt(x^2 + x)) =>
x / (x + sqrt(x^2 + x)) =>
x / (x + x * sqrt(1 + 1/x)) =>
1 / (1 + sqrt(1 + (1/x)))
x goes to infinity
1 / (1 + sqrt(1 + 1/inf)) =>
1 / (1 + sqrt(1 + 0)) =>
1 / (1 + sqrt(1)) =>
1 / (1 + 1) =>
1 / 2
- Anonymous5 years ago
limit x-> infinity of x*[sqrt(1 + (1/x)) - 1]
limit x->infinity [sqrt(1 + (1/x)) - 1]/(1/x) as now you have indeterminate form
use l'hopital rule {d/dx [sqrt(1 + (1/x)) - 1]}/(d/dx 1/x)
