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What is the limit of the sequence 3n²/(n²−3) as n approaches infinity?

The limit of the sequence 3n2n2−3\frac{3n^2}{n^2-3} as nn approaches infinity is 3. This is determined by observing that as nn grows large, the constant −3-3 in the denominator becomes negligible compared to n2n^2, so the expression behaves like 3n2n2=3\frac{3n^2}{n^2} = 3. The video visualizes this convergence with a horizontal asymptote at y=3y=3.

Conditions

  • Sequence is an=3n2n2−3a_n = \frac{3n^2}{n^2-3}.
  • n→∞n \to \infty.

Reasoning, step by step

  1. Identify the highest power of nn in the numerator and denominator.
  2. Divide both numerator and denominator by n2n^2.
  3. Observe that 31−3/n2\frac{3}{1 - 3/n^2} approaches 31−0=3\frac{3}{1 - 0} = 3.
  4. Confirm the horizontal asymptote at y=3y=3 in the visualization.

Example

The video presents the limit statement: lim⁡(n→∞)3n\lim (n\to \infty ) 3n²/(n² - 3) = 3. A coordinate system plots discrete points against a horizontal asymptote at y=3y=3.

Common misconceptions

  • Believing the limit is undefined because of the −3-3.
  • Thinking the limit is 0 because the denominator grows.

Watch the explanation

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.