L’Hôpital’s rule, question 14

Topic: L’Hôpital’s rule
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §6 CLP-1 §3.7Key results

Evaluate the limit.

\lim_{x \to \infty} \left(e^{x} + x\right)^{1/x}

Your answer

An exact expression, or a decimal to two decimal places.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §6 CLP-1 §3.7Key results

Worked solution

Strategy

Take the logarithm to turn the power into a quotient, settle that quotient with the rule and the fact that eˣ outgrows x, then exponentiate.

Working

$e^{x} + x \to \infty, \qquad \frac{1}{x} \to 0^{+} \qquad \bigl(\text{form } \infty^{0}\bigr)$
Indeterminate: a growing base against a vanishing exponent.
$\lim_{x \to \infty} \log\left(\left(e^{x} + x\right)^{1/x}\right) = \lim_{x \to \infty} \frac{\log(e^{x} + x)}{x} \;\overset{\infty/\infty}{=}\; \lim_{x \to \infty} \frac{e^{x} + 1}{e^{x} + x}$
Bring the exponent down, then one application; the chain rule gives the derivative of log(eˣ + x).
$= \lim_{x \to \infty} \frac{1 + e^{-x}}{1 + x e^{-x}} = \frac{1 + 0}{1 + 0} = 1$
Still ∞/∞, so divide through by eˣ; x/eˣ → 0 by one more application of the rule (1/eˣ → 0).
$\lim_{x \to \infty} \left(e^{x} + x\right)^{1/x} = \lim_{x \to \infty} e^{\log\left(\left(e^{x} + x\right)^{1/x}\right)} = e^{1} = e \approx 2.72$
The expression is e to the power of its own logarithm, and eᵘ is continuous, so the limit passes into the exponent.

Answer

e

Check

At x = 10 the power is about 2.7184, close to e ≈ 2.7183, because the added x is tiny beside eˣ.

Where this goes wrong

Answering 1 because the exponent tends to 0, or +∞ because the base grows. In an ∞⁰ form the two compete; the logarithm turns the power into a quotient that the rule can settle.

Leans on

Comparing exponentials and powers, from Power functions and first sketches

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