Limits, question 12: UBC MATH 100 practice

Question 12 of 30Standard3 marks · 5 minTheory UBC Math 100 §2.2CLP-1 §1.5Key results

Which list gives every horizontal and vertical asymptote of the graph of the function below?

f(x) = \frac{3e^{x} + 2}{e^{x} - 1}

Horizontal: y = 3 as x → ∞, and y = −2 as x → −∞. Vertical: x = 0.
Horizontal: y = 3 only, at both ends. Vertical: x = 0.
Horizontal: y = 3 as x → ∞, and y = −2 as x → −∞. Vertical: x = 1.
Horizontal: y = 3 only, at both ends. Vertical: x = 1.

How sure are you?

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Theory UBC Math 100 §2.2 CLP-1 §1.5Key results

Worked solution

Strategy

An exponential behaves differently at the two ends, so find the limit at ∞ and at −∞ separately. Then find where the denominator vanishes and the numerator does not.

Working

$x \to \infty:\quad \frac{3e^{x} + 2}{e^{x} - 1} = \frac{3 + 2e^{-x}}{1 - e^{-x}} \to \frac{3 + 0}{1 - 0} = 3$
Divide top and bottom by e^x; the terms in e^−x vanish.
$x \to -\infty:\quad e^{x} \to 0, \qquad \frac{3e^{x} + 2}{e^{x} - 1} \to \frac{0 + 2}{0 - 1} = -2$
On the far left e^x shrinks to 0, so only the constants survive.
$e^{x} - 1 = 0 \iff e^{x} = 1 \iff x = 0, \qquad 3e^{0} + 2 = 5 \neq 0$
The denominator vanishes only at x = 0, and the numerator does not, so the function blows up there.
$\lim_{x \to 0^{+}} f(x) = +\infty, \qquad \lim_{x \to 0^{-}} f(x) = -\infty$
e^x − 1 has the same sign as x, which settles each side.

Answer

y = 3 \ (x \to \infty), \quad y = -2 \ (x \to -\infty), \quad x = 0

Check

f(10) ≈ 3.0002, f(−10) ≈ −2.0002 and f(0.01) ≈ 500, so the values settle at 3 on the right and −2 on the left, and blow up at 0.

Where this goes wrong

Giving one horizontal asymptote for both ends, as for a rational function. e^x is huge on the right and nearly 0 on the left, so the two ends settle at different heights.

The idea behind it

The limits lesson, a figure to move, in a few steps

Leans on

The graph of the exponential function, from Functions, graphs and inverses

Try a similar question

Find the horizontal asymptote of the graph of the function below as

x \to -\infty

: enter the value of

y

it approaches.

f(x) = \frac{\sqrt{9x^{2} + 4}}{2x - 1}

My working

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Limits practice question 12