L’Hôpital’s rule, question 25: UBC MATH 100 practice

Question 25 of 27Challenging3 marks · 5 minTheory UBC Math 100 §6.2CLP-1 §3.7Key results

Two students start on the limit below. Student P applies l’Hôpital’s rule, and student Q divides the top and the bottom by $3^{x}$ . Which statement is right?

\lim_{x \to \infty} \frac{4\cdot 3^{x} + 2^{x}}{3^{x} + 5\cdot 2^{x}}

Both finish with 4: after one application of the rule, the factors log 3 and log 2 cancel.
Only Q finishes, and the limit is 4: each application of the rule gives another ∞/∞ quotient of the same shape.
Neither finishes, so the limit does not exist: the rule gives ∞/∞ every time.
Only Q finishes, and the limit is 1/5: for large x the 2^x terms decide.

How sure are you?

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Theory UBC Math 100 §6.2 CLP-1 §3.7Key results

Worked solution

Strategy

Think before differentiating. The rule cannot finish here, because differentiating an exponential only multiplies it by a constant; dividing by the fastest-growing term finishes in a line.

Working

$\overset{\text{l'H}}{\to}\ \frac{4\log 3\cdot 3^{x} + \log 2\cdot 2^{x}}{\log 3\cdot 3^{x} + 5\log 2\cdot 2^{x}}$
One application, using (a^x)′ = a^x log a: still ∞/∞ and of the same shape. Every further application multiplies the 3^x terms by log 3 and the 2^x terms by log 2 again, so the rule never reaches a form it can settle.
$\frac{4\cdot 3^{x} + 2^{x}}{3^{x} + 5\cdot 2^{x}} = \frac{4 + \left(\tfrac{2}{3}\right)^{x}}{1 + 5\left(\tfrac{2}{3}\right)^{x}}$
Divide every term by 3^x, the fastest-growing term: 2^x/3^x = (2/3)^x.
$\left(\tfrac{2}{3}\right)^{x} \to 0 \quad\Longrightarrow\quad \lim_{x \to \infty} \frac{4 + \left(\tfrac{2}{3}\right)^{x}}{1 + 5\left(\tfrac{2}{3}\right)^{x}} = \frac{4 + 0}{1 + 0} = 4$
A base between 0 and 1 raised to a growing power tends to 0.

Answer

\text{Only Q finishes; the limit is } 4

Check

At x = 10 the quotient is about 3.70, at x = 20 about 3.994 and at x = 30 about 3.9999: it creeps up to 4 as (2/3)^x fades.

Where this goes wrong

Applying the rule again and again in the hope that it settles, or deciding that the limit does not exist when it never does. When every term is an exponential, ask which grows fastest and divide by it.

The idea behind it

The L’Hôpital’s rule lesson, a figure to move, in a few steps

Leans on

Limits at infinity, from Limits

Try a similar question

Evaluate the limit.

\lim_{x \to \infty} \frac{7x + \log x}{2x - \log x}

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L’Hôpital’s rule practice question 25