Math · Calculus I · Worked example

Compare the growth of x² and eˣ

Evaluate the limit of x²/eˣ as x → ∞.

lim_{x \to \infty} \frac{x^{2}}{e^{x}}

Check the form

As x → ∞, both x² and eˣ grow without bound, so the form is ∞/∞.

Apply the rule

Differentiate the top and bottom: 2x over eˣ. The form is still ∞/∞.

lim_{x \to \infty} \frac{x^{2}}{e^{x}} = lim_{x \to \infty} \frac{2 x}{e^{x}}

Apply it again

Differentiating again gives 2/eˣ. The numerator is now constant while the denominator grows without bound, so the quotient tends to 0.

lim_{x \to \infty} \frac{2 x}{e^{x}} = lim_{x \to \infty} \frac{2}{e^{x}} = 0

Read the result

The exponential eventually dwarfs the square: at x = 10, x²/eˣ ≈ 0.0045, and at x = 20 it is about 8.2 × 10⁻⁷. The same argument, used n times, shows xⁿ/eˣ → 0 for every power n.

Result

The limit is 0: eˣ grows faster than x².

lim_{x \to \infty} \frac{x^{2}}{e^{x}} = 0

Your turn

Evaluate the limit of (ln x)/x as x → ∞.

Show the answer and explanation

0.

The form is ∞/∞. Differentiating the top and bottom gives (1/x)/1 = 1/x, which tends to 0: logarithms grow more slowly than x.

lim_{x \to \infty} \frac{ln x}{x} = lim_{x \to \infty} \frac{1}{x} = 0

Keep exploring

Open the graph of x²/eˣ: it peaks at x = 2, where its value is 4/e² ≈ 0.54, then falls toward 0.

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Sources and scope

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OpenStax Calculus Volume 1 — L’Hôpital’s rule

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Graph x²/eˣ in Graph Open worked example on a board Limit of a quotient in Math Reference

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