Practice / Calculus

Evaluating limits: factoring, conjugates and limits at infinity

Ten limits evaluated exactly: 0/0 forms cleared by factoring, by a difference of cubes and by the conjugate, the difference quotient behind every derivative, sin 3x / x, one-sided limits that disagree, and limits at ±∞ including √(x²) = |x| and ∞ − ∞, with worked solutions and the mistakes that read 0/0 as an answer.

Before you start

A limit asks what value $f(x)$ approaches as $x$ approaches $a$ not what $f(a)$ is, and every derivative is one of them. When substituting $a$ gives a number, that number is the limit; when it gives $\tfrac00$ or $\tfrac\infty\infty$ the work is to rewrite $f$ until substitution does give a number. These ten problems run from factoring and the conjugate to the difference quotient, one-sided limits, and limits at infinity, where the sign of $\sqrt{x^2}$ and the form $\infty - \infty$ catch people out.

$\lim_{x\to a} f(x) = L$ means $f(x)$ gets as close to $L$ as you like for every $x$ close enough to $a$ with $x \neq a$ The value $f(a)$ or whether $f(a)$ is defined at all, plays no part.
If $f$ is built from polynomials, roots, exponentials, logs and trig functions by $+$ $-$ $\times$ $\div$ and composition, and $f(a)$ is defined, then $\lim_{x\to a} f(x) = f(a)$ substitute.
$\tfrac00$ is not a value. It says the limit depends on how fast the top and bottom go to 0, and the expression has to be rewritten before you substitute. Two expressions that agree for every $x \neq a$ have the same limit at $a$ which is what makes canceling a common factor $(x - a)$ legal.
One-sided limits: $x \to a^-$ approaches from below, $x \to a^+$ from above. The two-sided limit exists exactly when both one-sided limits exist and are equal.
Standard limit: $\lim_{t\to0} \dfrac{\sin t}{t} = 1$ with $t$ in radians.
At $\pm\infty$ divide the top and bottom by the highest power of $x$ in the denominator, and use $\tfrac{1}{x^n} \to 0$ for $n > 0$ For a root, $\sqrt{x^2} = |x|$ which is $-x$ when $x < 0$

Problems

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Find $\displaystyle\lim_{x\to3} \frac{x^2 - 9}{x - 3}$
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Find $\displaystyle\lim_{x\to2} \frac{x^2 + x - 6}{x^2 - 4}$
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Find $\displaystyle\lim_{x\to0} \frac{\sqrt{x + 4} - 2}{x}$
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Find $\displaystyle\lim_{x\to1} \frac{x^3 - 1}{x^2 - 1}$
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Find $\displaystyle\lim_{h\to0} \frac{(2 + h)^2 - 4}{h}$ and say which derivative it is.
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Find $\displaystyle\lim_{x\to\infty} \frac{3x^2 - x + 5}{2x^2 + 7}$
···
Find $\displaystyle\lim_{x\to-\infty} \frac{\sqrt{4x^2 + 1}}{x + 3}$
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Find $\displaystyle\lim_{x\to2^-} \frac{|x - 2|}{x - 2}$ and $\displaystyle\lim_{x\to2^+} \frac{|x - 2|}{x - 2}$ Does $\displaystyle\lim_{x\to2} \frac{|x - 2|}{x - 2}$ exist?
···
Find $\displaystyle\lim_{x\to0} \frac{\sin 3x}{x}$
···
Find $\displaystyle\lim_{x\to\infty} \left(\sqrt{x^2 + 6x} - x\right)$

Answers

$6$
$\tfrac54$
$\tfrac14$
$\tfrac32$
$4$ it is $f'(2)$ for $f(x) = x^2$
$\tfrac32$
$-2$
$-1$ from the left, $1$ from the right; the two-sided limit does not exist
$3$
$3$

Worked solutions

Problem 1

Find $\displaystyle\lim_{x\to3} \frac{x^2 - 9}{x - 3}$

Substituting $x = 3$ gives $\tfrac{9 - 9}{3 - 3} = \tfrac00$ Not an answer: the expression has to be rewritten first.
$\dfrac{x^2 - 9}{x - 3} = \dfrac{(x - 3)(x + 3)}{x - 3} = x + 3$ for $x \neq 3$ Difference of squares. The factor $x - 3$ is the one that made both the top and the bottom 0, and canceling it is legal because the limit never uses $x = 3$ itself.
$\displaystyle\lim_{x\to3}(x + 3) = 6$ $x + 3$ is a polynomial, so substitute.
$6$ Steps 2 and 3.

Problem 2

Find $\displaystyle\lim_{x\to2} \frac{x^2 + x - 6}{x^2 - 4}$

Substituting $x = 2$ gives $\tfrac{4 + 2 - 6}{4 - 4} = \tfrac00$ Both polynomials are 0 at 2, so both have the factor $x - 2$
$x^2 + x - 6 = (x - 2)(x + 3)$ and $x^2 - 4 = (x - 2)(x + 2)$ Factor both. In the first, $-2$ and $3$ multiply to $-6$ and add to $1$
$\dfrac{x^2 + x - 6}{x^2 - 4} = \dfrac{x + 3}{x + 2}$ for $x \neq 2$ Cancel the common factor $x - 2$
$\tfrac54$ Substitute $x = 2$ into $\tfrac{x + 3}{x + 2}$ which is defined there: $\tfrac{5}{4}$

Problem 3

Find $\displaystyle\lim_{x\to0} \frac{\sqrt{x + 4} - 2}{x}$

Substituting $x = 0$ gives $\tfrac{2 - 2}{0} = \tfrac00$ There is no polynomial to factor, so use the conjugate of the root expression.
$\dfrac{\sqrt{x + 4} - 2}{x}\cdot\dfrac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}$ Multiplying by $\tfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}$ which is 1, changes the form and not the value. The top and the bottom both get the factor.
$(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2) = (x + 4) - 4 = x$ Difference of squares, $(p - q)(p + q) = p^2 - q^2$ the root squares away.
$\dfrac{x}{x(\sqrt{x + 4} + 2)} = \dfrac{1}{\sqrt{x + 4} + 2}$ for $x \neq 0$ Step 3 on top, the bottom left in factored form so the $x$ can cancel.
$\tfrac14$ Substitute $x = 0$ $\tfrac{1}{\sqrt4 + 2} = \tfrac{1}{4}$

Problem 4

Find $\displaystyle\lim_{x\to1} \frac{x^3 - 1}{x^2 - 1}$

Substituting $x = 1$ gives $\tfrac00$ Both are 0 at $x = 1$ so both have the factor $x - 1$
$x^3 - 1 = (x - 1)(x^2 + x + 1)$ and $x^2 - 1 = (x - 1)(x + 1)$ Difference of cubes, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and difference of squares.
$\dfrac{x^3 - 1}{x^2 - 1} = \dfrac{x^2 + x + 1}{x + 1}$ for $x \neq 1$ Cancel $x - 1$
$\tfrac32$ Substitute $x = 1$ $\tfrac{1 + 1 + 1}{1 + 1} = \tfrac32$

Problem 5

Find $\displaystyle\lim_{h\to0} \frac{(2 + h)^2 - 4}{h}$ and say which derivative it is.

$(2 + h)^2 - 4 = 4 + 4h + h^2 - 4 = 4h + h^2$ Expand the square, middle term included. Substituting $h = 0$ in the original gives $\tfrac00$ so it has to be rewritten.
$\dfrac{4h + h^2}{h} = 4 + h$ for $h \neq 0$ Every term on top has a factor $h$ so it cancels with the bottom.
$\displaystyle\lim_{h\to0}(4 + h) = 4$ Substitute $h = 0$
$4$ it is $f'(2)$ for $f(x) = x^2$ The derivative is defined as $f'(a) = \lim_{h\to0}\tfrac{f(a + h) - f(a)}{h}$ and here $f(a + h) = (2 + h)^2$ $f(a) = 4$ It agrees with $(x^2)' = 2x$ at $x = 2$

Problem 6

Find $\displaystyle\lim_{x\to\infty} \frac{3x^2 - x + 5}{2x^2 + 7}$

Substituting is not possible, and the top and bottom both grow without bound: the form is $\tfrac\infty\infty$ Rewrite so that every term has a limit.
$\dfrac{3x^2 - x + 5}{2x^2 + 7} = \dfrac{3 - \tfrac1x + \tfrac{5}{x^2}}{2 + \tfrac{7}{x^2}}$ Divide every term, top and bottom, by $x^2$ the highest power in the denominator. This is multiplying by $\tfrac{1/x^2}{1/x^2} = 1$
$\tfrac1x \to 0$ $\tfrac{5}{x^2} \to 0$ and $\tfrac{7}{x^2} \to 0$ as $x \to \infty$ A fixed number divided by an ever larger one.
$\tfrac32$ The limit of the top is 3 and of the bottom is 2. When the top and bottom have the same degree, the limit is the ratio of the leading coefficients.

Problem 7

Find $\displaystyle\lim_{x\to-\infty} \frac{\sqrt{4x^2 + 1}}{x + 3}$

$\sqrt{4x^2 + 1} = \sqrt{x^2}\,\sqrt{4 + \tfrac{1}{x^2}} = |x|\sqrt{4 + \tfrac{1}{x^2}}$ Factor $x^2$ out from under the root. $\sqrt{x^2}$ is $|x|$ not $x$ the root is never negative.
$|x| = -x$ for $x < 0$ $x \to -\infty$ so from some point on $x$ is negative.
$\dfrac{\sqrt{4x^2 + 1}}{x + 3} = \dfrac{-x\sqrt{4 + \tfrac{1}{x^2}}}{x\big(1 + \tfrac3x\big)} = \dfrac{-\sqrt{4 + \tfrac{1}{x^2}}}{1 + \tfrac3x}$ Steps 1 and 2 on top, $x$ factored out of the bottom, then cancel $x$
$-2$ $\tfrac{1}{x^2} \to 0$ and $\tfrac3x \to 0$ so the limit is $\tfrac{-\sqrt4}{1} = -2$ Sanity check: the top is positive and the bottom is negative for large negative $x$ so the limit has to be negative.

Problem 8

Find $\displaystyle\lim_{x\to2^-} \frac{|x - 2|}{x - 2}$ and $\displaystyle\lim_{x\to2^+} \frac{|x - 2|}{x - 2}$ Does $\displaystyle\lim_{x\to2} \frac{|x - 2|}{x - 2}$ exist?

For $x < 2$ $|x - 2| = -(x - 2)$ so $\dfrac{|x - 2|}{x - 2} = -1$ Below 2, $x - 2$ is negative and the absolute value flips its sign.
$\displaystyle\lim_{x\to2^-} \frac{|x - 2|}{x - 2} = -1$ The expression is the constant $-1$ for every $x$ just below 2.
For $x > 2$ $|x - 2| = x - 2$ so the expression is $1$ and $\displaystyle\lim_{x\to2^+} = 1$ Above 2, $x - 2$ is positive and the absolute value changes nothing.
$-1$ from the left, $1$ from the right; the two-sided limit does not existThe one-sided limits exist but differ, so there is no single value that $f(x)$ approaches from both sides.

Problem 9

Find $\displaystyle\lim_{x\to0} \frac{\sin 3x}{x}$

Substituting $x = 0$ gives $\tfrac{\sin 0}{0} = \tfrac00$ The standard limit $\tfrac{\sin t}{t} \to 1$ applies only when the angle and the denominator are the same quantity.
$\dfrac{\sin 3x}{x} = 3\cdot\dfrac{\sin 3x}{3x}$ Multiply the top and the bottom by 3, so the denominator matches the angle.
$t = 3x \to 0$ as $x \to 0$ so $\dfrac{\sin 3x}{3x} = \dfrac{\sin t}{t} \to 1$ The substitution $t = 3x$ turns the expression into the standard limit.
$3$ $3 \cdot 1$ The same argument gives $\lim_{x\to0}\tfrac{\sin kx}{x} = k$ for any constant $k$

Problem 10

Find $\displaystyle\lim_{x\to\infty} \left(\sqrt{x^2 + 6x} - x\right)$

Both $\sqrt{x^2 + 6x}$ and $x$ go to $\infty$ the form is $\infty - \infty$ Like $\tfrac00$ that is not a value. Multiply by the conjugate to turn the difference into a quotient.
$\left(\sqrt{x^2 + 6x} - x\right)\dfrac{\sqrt{x^2 + 6x} + x}{\sqrt{x^2 + 6x} + x} = \dfrac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x}$ Difference of squares on top: the $x^2$ terms cancel.
$\dfrac{6x}{\sqrt{x^2 + 6x} + x} = \dfrac{6x}{x\sqrt{1 + \tfrac6x} + x} = \dfrac{6}{\sqrt{1 + \tfrac6x} + 1}$ $\sqrt{x^2 + 6x} = |x|\sqrt{1 + \tfrac6x}$ and $|x| = x$ for $x > 0$ then cancel $x$
$3$ $\tfrac6x \to 0$ so the limit is $\tfrac{6}{1 + 1} = 3$ The root is about $x + 3$ for large $x$ since $(x + 3)^2 = x^2 + 6x + 9$

Where this goes wrong

1. Reading 0/0 as the answer

Substitution is the first thing to try, and when it produces $\tfrac00$ it looks like a result.

$\displaystyle\lim_{x\to3}\frac{x^2 - 9}{x - 3}$ substituting gives $\tfrac00$ Right so far: Problem 1, step 1.
“Division by zero is undefined, so the limit is undefined.”The habit that causes the mistake: treating the limit as the value at $x = 3$
$\displaystyle\lim_{x\to3}\frac{x^2 - 9}{x - 3}$ does not existThe function is undefined at 3, but the limit never uses $x = 3$ For every other $x$ it equals $x + 3$ which approaches 6; at $x = 3.001$ the expression is $6.001$ $\tfrac00$ is a signal to rewrite, never a final answer, and neither is "0" or "1".

2. Canceling terms instead of factors

The $x^2$ on top and the $x^2$ on the bottom look like they should cancel.

$\displaystyle\lim_{x\to2}\frac{x^2 + x - 6}{x^2 - 4}$ Right so far: Problem 2.
“Cancel the $x^2$ from the top and the bottom.”The habit that causes the mistake: canceling a term that is added, not multiplied.
$\dfrac{x^2 + x - 6}{x^2 - 4} = \dfrac{x - 6}{-4}$ so the limit is $\tfrac{-4}{-4} = 1$ Only a factor of the whole top and the whole bottom cancels. At $x = 3$ the left side is $\tfrac65$ and the right side is $\tfrac34$ so the two are not equal. Factor first: the common factor is $x - 2$ and the limit is $\tfrac54$ (Problem 2).

3. Multiplying only the top by the conjugate

The conjugate is there to remove the root from the top, so the bottom gets left alone.

$\displaystyle\lim_{x\to0}\frac{\sqrt{x + 4} - 2}{x}$ using the conjugate $\sqrt{x + 4} + 2$ Right so far: Problem 3.
“ $(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2) = x$ so the numerator becomes $x$ ”The numerator is right; the bottom has not been touched.
$\dfrac{\sqrt{x + 4} - 2}{x} = \dfrac{x}{x} = 1$ Multiplying only the top by $\sqrt{x+4} + 2$ multiplies the whole expression by a number near 4, which changes it. The bottom must become $x(\sqrt{x + 4} + 2)$ the answer is $\tfrac14$ not 1.

4. Writing √(x²) = x when x → −∞

Pulling $x$ out of a square root is a reflex, and for positive $x$ it is right.

$\displaystyle\lim_{x\to-\infty}\frac{\sqrt{4x^2 + 1}}{x + 3}$ Right so far: Problem 7.
“ $\sqrt{4x^2 + 1} = x\sqrt{4 + 1/x^2}$ ”The habit that causes the mistake: $\sqrt{x^2} = x$ which holds only for $x \ge 0$
$\dfrac{x\sqrt{4 + 1/x^2}}{x(1 + 3/x)} \to 2$ $\sqrt{x^2} = |x| = -x$ for negative $x$ so a minus sign is missing. At $x = -1000$ the expression is about $\tfrac{2000}{-997} \approx -2.006$ a positive root over a negative denominator. The limit is $-2$

5. Treating ∞ − ∞ as 0

For large $x$ $\sqrt{x^2 + 6x}$ and $x$ are both huge and nearly equal, so their difference looks negligible.

$\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2 + 6x} - x\right)$ Right so far: Problem 10.
“For large $x$ $x^2 + 6x \approx x^2$ so $\sqrt{x^2 + 6x} \approx x$ ”True as a ratio, $\sqrt{x^2 + 6x}/x \to 1$ but a ratio near 1 says nothing about the difference.
$\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2 + 6x} - x\right) = \infty - \infty = 0$ At $x = 10^6$ the difference is $2.999995\ldots$ nowhere near 0. The $6x$ that was dropped is exactly what the difference measures. Use the conjugate (Problem 10): the limit is 3.

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