Limits, question 29

Topic: Limits
Difficulty: Capstone
Suggested time: 9 min
Marks: 5
Theory: UBC Math 100 §2.1.2 CLP-1 §1.5Key results

For exactly one pair of constants $a$ and $b$ , the limit below exists and equals 2. Find $a$ and $b$ , and enter $a + b$ .

\lim_{x \to -\infty} \Bigl(\sqrt{36x^{2} + ax + 5} + bx\Bigr) = 2

Your answer

A number.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §2.1.2 CLP-1 §1.5Key results

Worked solution

Strategy

Two conditions fix two constants. The limit is finite only if bx cancels the growth of the root, which fixes b; then the conjugate turns the ∞ − ∞ form into a quotient whose limit, set equal to 2, fixes a. Both steps turn on the sign of √(x²) for negative x.

Working

$x < 0:\quad \sqrt{36x^{2} + ax + 5} = |x|\sqrt{36 + \tfrac{a}{x} + \tfrac{5}{x^{2}}} = -x\sqrt{36 + \tfrac{a}{x} + \tfrac{5}{x^{2}}}$
For negative x, √(x²) = |x| = −x.
$\sqrt{36x^{2} + ax + 5} + bx = x\left(b - \sqrt{36 + \tfrac{a}{x} + \tfrac{5}{x^{2}}}\right)$
The bracket tends to b − 6 while x → −∞, so the expression is unbounded unless b = 6.
$b = 6:\quad \sqrt{36x^{2} + ax + 5} + 6x = \frac{(36x^{2} + ax + 5) - 36x^{2}}{\sqrt{36x^{2} + ax + 5} - 6x} = \frac{ax + 5}{\sqrt{36x^{2} + ax + 5} - 6x}$
Multiply and divide by the conjugate; the x² terms cancel.
$= \frac{ax + 5}{-x\left(\sqrt{36 + \frac{a}{x} + \frac{5}{x^{2}}} + 6\right)} = \frac{a + \frac{5}{x}}{-\left(\sqrt{36 + \frac{a}{x} + \frac{5}{x^{2}}} + 6\right)} \to -\frac{a}{12}$
Use the same −x for the root in the denominator, then divide the top and the bottom by x.
$-\frac{a}{12} = 2 \quad\Longrightarrow\quad a = -24, \qquad a + b = -24 + 6 = -18$
Set the limit equal to 2.

Answer

a = -24, \quad b = 6, \qquad a + b = -18

Check

With a = −24 and b = 6, at x = −1000 the expression is √36,024,005 − 6000 ≈ 2.00008, close to 2.

Where this goes wrong

Writing √(36x²) = 6x. For negative x it is −6x, so the b that cancels the growth is 6, not −6, and the same sign decides the denominator after the conjugate. With the sign wrong in both places the working gives b = −6 and a = 24, a pair for which the expression tends to +∞.

Leans on

Square roots and absolute value, from Functions, graphs and inverses

Try a similar question

Evaluate the limit.

\lim_{x \to -\infty} \Bigl(\sqrt{x^{2} + 5x} - \sqrt{x^{2} - 3x}\Bigr)

My working

Saved in this browser only.

Limits practice question 29