Limits, question 25

Topic: Limits
Difficulty: Challenging
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §2.2 CLP-1 §1.4Key results

Which statement about the vertical asymptotes of the graph of the function below is correct?

g(x) = \frac{x^{2} - 4}{\sqrt{x + 7} - 3}

There is no vertical asymptote: the graph has a hole at x = 2, where g(x) tends to 24.
x = 2 is a vertical asymptote, because the denominator is 0 there.
There is no vertical asymptote: the graph has a hole at x = 2, where g(x) tends to 4.
x = −7 is a vertical asymptote, because the domain ends there.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §2.2 CLP-1 §1.4Key results

Worked solution

Strategy

A zero of the denominator gives an asymptote only if the quotient blows up there. At x = 2 the numerator vanishes too, so rationalise the denominator and see what is left. Check the edge of the domain as well.

Working

$\sqrt{x + 7} - 3 = 0 \iff x + 7 = 9 \iff x = 2, \qquad 2^{2} - 4 = 0$
Both parts vanish at 2: a 0/0 form, not yet a verdict.
$g(x) = \frac{(x^{2} - 4)\bigl(\sqrt{x + 7} + 3\bigr)}{(x + 7) - 9} = \frac{(x - 2)(x + 2)\bigl(\sqrt{x + 7} + 3\bigr)}{x - 2}$
Multiply the top and the bottom by the conjugate; the bottom becomes x − 2.
$\lim_{x \to 2} g(x) = \lim_{x \to 2} (x + 2)\bigl(\sqrt{x + 7} + 3\bigr) = 4 \cdot 6 = 24$
Cancel x − 2. The limit is finite, so x = 2 is a hole, not an asymptote.
$g(-7) = \frac{49 - 4}{0 - 3} = -15$
Where the domain begins, g has a finite value, so x = −7 is no asymptote either.

Answer

\text{no vertical asymptote; a hole at } x = 2, \text{ where } g(x) \to 24

Check

g(1.99) ≈ 23.93 and g(2.01) ≈ 24.07, so g settles near 24 instead of blowing up.

Where this goes wrong

Calling every zero of the denominator a vertical asymptote. When the numerator is 0 there too, cancel first: here the quotient tends to 24, so the graph has a hole.

Leans on

Rationalising with a conjugate, from Functions, graphs and inverses

Try a similar question

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

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

My working

Saved in this browser only.

Limits practice question 25