Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# Limit problem (high school calculus) help?

### 3 Answers

Relevance
• 1 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.

• Login to reply the answers
• Vaman
Lv 7
1 month ago

When x is very large, then you can neglect x. The equation becomes

3x^2/x=3x. The limit becomes infinity.

• Login to reply the answers
• 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 =>

inf

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.

Still have questions? Get your answers by asking now.