L’Hôpital’s rule, question 15

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

A deposit of $2000 earns interest at 5 % per year, compounded $n$ times a year, so after 10 years the balance in dollars is $B(n)$ below. Find the balance under continuous compounding, the limit of $B(n)$ as $n \to \infty$ .

B(n) = 2000\left(1 + \frac{0.05}{n}\right)^{10n}

Give the answer in dollars.

Your answer

An exact expression, or an amount to the nearest cent.

How sure are you?

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

Worked solution

Strategy

Continuous compounding is a 1^∞ limit. Take the logarithm of the growth factor, rewrite it as a 0/0 quotient, apply the rule, and exponentiate.

Working

$\lim_{n \to \infty} B(n) = 2000\lim_{n \to \infty} \left(1 + \frac{0.05}{n}\right)^{10n} \qquad \bigl(\text{form } 1^{\infty}\bigr)$
The deposit is a constant factor; the growth factor is the indeterminate part.
$\lim_{n \to \infty} \log\left[\left(1 + \frac{0.05}{n}\right)^{10n}\right] = \lim_{n \to \infty} 10n\log\!\left(1 + \frac{0.05}{n}\right) = \lim_{n \to \infty} \frac{10\log(1 + 0.05/n)}{1/n} \qquad \left(\tfrac{0}{0}\right)$
The logarithm brings the exponent down; then rewrite ∞·0 as 0/0, treating n as a continuous variable.
$\overset{\text{l'H}}{=} \lim_{n \to \infty} \frac{10\cdot\dfrac{-0.05/n^{2}}{1 + 0.05/n}}{-1/n^{2}} = \lim_{n \to \infty} \frac{0.5}{1 + 0.05/n} = 0.5$
The −1/n² factors cancel.
$\lim_{n \to \infty} B(n) = 2000\lim_{n \to \infty} e^{\log\left[\left(1 + 0.05/n\right)^{10n}\right]} = 2000e^{0.5} = 2000\sqrt{e} \approx 3297.44$
eᵘ is continuous, so the limit passes into the exponent; then multiply by the deposit.

Answer

2000e^{1/2} \approx \$3297.44

Check

Monthly compounding gives about $3294.02 and daily about $3297.33, creeping up toward the continuous $3297.44.

Where this goes wrong

Answering $2000 because the base tends to 1, or an unlimited balance because the exponent grows. Neither part decides a 1^∞ form; the logarithm shows the growth factor is e^{0.05 × 10} = e^0.5.

Try a similar question

Evaluate the limit.

\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{x}

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L’Hôpital’s rule practice question 15