Limits, question 23: UBC MATH 100 practice

Question 23 of 30Challenging3 marks · 6 minTheory UBC Math 100 §2.2CLP-1 §1.5Key results

The function below has a horizontal asymptote as $x \to \infty$ and another as $x \to -\infty$ . Find the $x$ -coordinate of the point where its graph crosses one of these asymptotes, or type none if it never does.

f(x) = \frac{4x + 3}{\sqrt{x^{2} + 1}}

Your answer

An exact fraction, or a decimal to three decimal places.

How sure are you?

SureNot sureGuessing

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

Worked solution

Strategy

Find the horizontal asymptote at each end separately, minding the sign of √(x²) for negative x. Then solve f(x) = L for each asymptote, checking every candidate after squaring, since squaring cannot tell L from −L.

Working

$\lim_{x \to \infty} \frac{4x + 3}{\sqrt{x^{2} + 1}} = \lim_{x \to \infty} \frac{4 + 3/x}{\sqrt{1 + 1/x^{2}}} = 4$
For x > 0, √(x² + 1) = x√(1 + 1/x²), so divide the top and the bottom by x. The asymptote on the right is y = 4.
$\lim_{x \to -\infty} \frac{4x + 3}{\sqrt{x^{2} + 1}} = \lim_{x \to -\infty} \frac{4 + 3/x}{-\sqrt{1 + 1/x^{2}}} = -4$
For x < 0, √(x² + 1) = |x|√(1 + 1/x²) = −x√(1 + 1/x²), which brings a minus sign. The asymptote on the left is y = −4.
$4x + 3 = 4\sqrt{x^{2} + 1} \;\Longrightarrow\; 16x^{2} + 24x + 9 = 16x^{2} + 16 \;\Longrightarrow\; x = \frac{7}{24}$
Square both sides and the x² terms cancel. At this x, 4x + 3 = 25/6 is positive, matching the positive right side, so it is a genuine solution: the graph crosses y = 4.
$4x + 3 = -4\sqrt{x^{2} + 1}$
This needs 4x + 3 < 0, but squaring leads to the same candidate x = 7/24, where 4x + 3 > 0. So the graph never meets y = −4.
$f\left(\frac{7}{24}\right) = \frac{25/6}{25/24} = 4$
Substituting back confirms the crossing: √((7/24)² + 1) = 25/24.

Answer

x = \frac{7}{24}

Check

f(0) = 3 is below 4 and f(1) = 7/√2 ≈ 4.95 is above it, so the graph does pass through y = 4 between x = 0 and x = 1.

Where this goes wrong

Answering none because a graph supposedly never crosses its horizontal asymptote. The asymptote only says where f heads as x → ±∞; this graph climbs through y = 4 at x = 7/24, rises to 5, and only then settles back toward 4 from above.

The idea behind it

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

Leans on

Square roots and absolute value, 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 23