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Exercise 13

Avg difficulty 2.5

Consider the function g(x)=x2xg(x) = x^{2x}, which is defined for all x>0x \gt 0. Observe that lim⁡x→0+g(x)\lim_{x \to 0^+} g(x) is indeterminate due to its form of 000^0. (Think about how we know that 0k=00^k = 0 for all k>0k \gt 0, while b0=1b^0 = 1 for all b≠0b \ne 0, but that neither rule can apply to 000^0.)

  1. Let h(x)=ln⁡(g(x))h(x) = \ln(g(x)). Explain why h(x)=2xln⁡(x)h(x) = 2x \ln(x).

  2. Next, explain why it is equivalent to write h(x)=2ln⁡(x)1xh(x) = \frac{2\ln(x)}{\frac{1}{x}}.

  3. Use L'Hôpital's Rule and your work in (b) to compute lim⁡x→0+h(x)\lim_{x \to 0^+} h(x).

  4. Based on the value of lim⁡x→0+h(x)\lim_{x \to 0^+} h(x), determine lim⁡x→0+g(x)\lim_{x \to 0^+} g(x).

Boelkins, Active Calculus, §3.2 Using derivatives to evaluate limits, Exercise 13

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.