Limit problem (high school calculus) help?
- JohnathanLv 71 month ago
You'd have 3x^2 / √(x^2) if you deal with only the highest exponents of x. For all non-negative x, √(x^2) = x, so since x approaches infinity here you'd have 3x^2 / x, which simplifies to 3x. This increases forever without limit (or, rather, the limit is infinity), so the limit of the whole expression is infinity.
Because you have a limit of infinity, there would be no horizontal asymptote; for that to be the case the limit would have to approach a certain value.
- VamanLv 71 month ago
When x is very large, then you can neglect x. The equation becomes
3x^2/x=3x. The limit becomes infinity.
- 1 month ago
(3x^2 + 2x) / sqrt(x^2 - 2x) =>
x * (3x + 2) / (sqrt(x^2 * (1 - 2/x))) =>
x * (3x + 2) / (x * sqrt(1 - 2/x)) =>
(3x + 2) / sqrt(1 - 2/x)
x goes to infinity
(3 * inf + 2) / sqrt(1 - 2/inf) =>
inf / sqrt(1 - 0) =>
inf / sqrt(1) =>
inf / 1 =>
As x goes to infinity, the limit goes to infinity. What does it say about the existence of horizontal asymptotes? Beats me, since there aren't any as x approaches infinity.