Limits, question 27

Topic: Limits
Difficulty: Challenging
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §2.1 CLP-1 §1.3Key results

Two students evaluate the one-sided limit $\displaystyle\lim_{x \to -2^{+}} \frac{x^{2} + x - 2}{x^{2} + 4x + 4}$ as shown. Which solution is right, and why?

\begin{aligned} &\textbf{Solution A} \\ \text{A1:}\quad & \frac{x^{2} + x - 2}{x^{2} + 4x + 4} = \frac{(x + 2)(x - 1)}{(x + 2)^{2}} \\ & = \frac{x - 1}{x + 2} \\[4pt] \text{A2:}\quad & \text{at } x = -2.01:\ \frac{-3.01}{-0.01} = 301 \\[4pt] \text{A3:}\quad & \text{large and positive, so the limit is } +\infty \\[14pt] &\textbf{Solution B} \\ \text{B1:}\quad & \frac{(x + 2)(x - 1)}{(x + 2)^{2}} = \frac{x - 1}{x + 2}, \quad x \neq -2 \\[4pt] \text{B2:}\quad & x \to -2^{+}:\quad x - 1 \to -3, \quad x + 2 \to 0^{+} \\[4pt] \text{B3:}\quad & \text{negative over small positive,} \\ & \text{so the limit is } -\infty \end{aligned}

A is right; B ends negative, but the denominator (x + 2)² is never negative, so the quotient cannot tend to −∞.
B is right; A tests x = −2.01, which lies to the left of −2, but x → −2⁺ approaches from the right.
Both are right: each shows the quotient blows up next to x = −2, so the limit is infinite either way.
Neither is right: after cancelling, the quotient tends to −3/0, which is undefined, so the limit does not exist.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §2.1 CLP-1 §1.3Key results

Worked solution

Strategy

Check each line of each solution. The cancelling in line 1 is shared and correct, so the question is which side of −2 each solution looks at.

Working

$\frac{x^{2} + x - 2}{x^{2} + 4x + 4} = \frac{(x + 2)(x - 1)}{(x + 2)^{2}} = \frac{x - 1}{x + 2}, \quad x \neq -2$
A1 and B1 are right: the form is 0/0 at −2, and cancelling x + 2 is allowed because the limit never uses x = −2 itself.
$x \to -2^{+} \implies x > -2, \quad \text{e.g. } x = -1.99$
The plus sign means from the right: values a little greater than −2.
$-2.01 < -2$
A2 goes wrong here. −2.01 lies to the left of −2, so it tests the limit from the left, which is +∞, not this one.
$x = -1.99: \quad \frac{-2.99}{0.01} = -299$
A test point on the correct side agrees with B2: the top is near −3 and the bottom is small and positive.
$\lim_{x \to -2^{+}} \frac{x - 1}{x + 2} = -\infty$
B3 is right, and so is the whole of B.

Answer

\text{B is right:}\quad \lim_{x \to -2^{+}} \frac{x^{2} + x - 2}{x^{2} + 4x + 4} = -\infty

Check

Just right of −2 the original top (x + 2)(x − 1) is negative while the bottom (x + 2)² is positive, so the quotient is negative, which rules out +∞.

Where this goes wrong

Testing a point on the wrong side of a negative number. −2.01 looks like it is past −2, but it is smaller, so it lies to the left; from the right means values such as −1.99.

Try a similar question

Evaluate the one-sided limit.

\lim_{x \to 4^{+}} \frac{x + 2}{4 - x}

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Limits practice question 27