Limit Laws and Composite Limits

Limit laws

Suppose

\lim_{x \to a} f(x) = L

and

\lim_{x \to a} g(x) = M

, where

L

and

M

are numbers, and let

c

be a constant. Then

\begin{aligned} &\lim_{x \to a} \bigl[f(x) \pm g(x)\bigr] = L \pm M \\[6pt] &\lim_{x \to a} c\,f(x) = cL \\[6pt] &\lim_{x \to a} f(x)\,g(x) = LM \\[6pt] &\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}, \quad \text{if } M \ne 0. \end{aligned}

For a positive integer

n

\lim_{x \to a} [f(x)]^n = L^n

, and

\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}

, where

L > 0

n

is even. The same laws hold for one-sided limits.

Two limits come straight from the definition:

\lim_{x \to a} c = c

for a constant

c

, and

\lim_{x \to a} x = a

. Every polynomial is built from constants and

x

by sums and products. A rational function is a quotient of two polynomials. So the laws give the next result.

Example 1

Suppose

\lim_{x \to 3} f(x) = 4

and

\lim_{x \to 3} g(x) = -2

. Find each limit, or explain why the given information doesn't determine it.

\begin{aligned} &\text{(a)}\;\; \lim_{x \to 3} \bigl[2f(x) - 5g(x)\bigr] \\[10pt] &\text{(b)}\;\; \lim_{x \to 3} \bigl([f(x)]^2 + x\,g(x)\bigr) \\[10pt] &\text{(c)}\;\; \lim_{x \to 3} \frac{\sqrt{f(x)}}{g(x) + x} \\[10pt] &\text{(d)}\;\; \lim_{x \to 3} \frac{f(x) - 4}{g(x) + 2} \end{aligned}

(a) By the limit laws, the limit is

2(4) - 5(-2) = 18

(b) Here

\lim_{x \to 3} x = 3

, so the limit is

4^2 + 3(-2) = 10

(c) The numerator's limit is

\sqrt{4} = 2

, since

4 > 0

. The denominator's limit is

-2 + 3 = 1

, which isn't

0

, so the law for quotients applies:

\lim_{x \to 3} \frac{\sqrt{f(x)}}{g(x) + x} = \frac{\sqrt{4}}{1} = 2.

(d) The numerator and the denominator both approach

0

, so the law for quotients doesn't apply. The limit depends on more than the two given limits. For instance,

f(x) = x + 1

and

g(x) = x - 5

meet the conditions and give a quotient of

1

for

x \ne 3

. With

f(x) = 2x - 2

instead, the quotient is

2

. So the information doesn't determine this limit.

Limits from two graphs

Example 3

The graph of

f

from Limits from Graphs and Tables is shown with the graph of a second function

g

. Find each limit, or explain why it doesn't exist.

The graph of f on [−5, 7]. The dashed line x = 4 is a vertical asymptote.

The graph of g.

\begin{aligned} &\text{(a)}\;\; \lim_{x \to 1} \bigl[f(x) + g(x)\bigr] \\[10pt] &\text{(b)}\;\; \lim_{x \to 1} f(x)\,g(x) \\[10pt] &\text{(c)}\;\; \lim_{x \to 1} (x - 1)\,f(x) \end{aligned}

(a) Neither

f

nor

g

has a limit at

1

, so the law for sums doesn't apply. Work with one-sided limits instead. From the left,

f

approaches

2

and

g

approaches

1

. From the right,

f

approaches

-1

and

g

approaches

4

. The one-sided limits of the sum are

\begin{aligned} &\lim_{x \to 1^-} \bigl[f(x) + g(x)\bigr] = 2 + 1 = 3, \\[4pt] &\lim_{x \to 1^+} \bigl[f(x) + g(x)\bigr] = -1 + 4 = 3. \end{aligned}

They agree, so

\lim_{x \to 1} [f(x) + g(x)] = 3

. The two jumps cancel.

(b) From the left the product approaches

2 \cdot 1 = 2

, and from the right it approaches

(-1)(4) = -4

. The one-sided limits differ, so the limit doesn't exist.

(c) Again

f

has no limit at

1

, but each one-sided limit exists, and

x - 1

approaches

0

from both sides:

\begin{aligned} &\lim_{x \to 1^-} (x - 1)\,f(x) = 0 \cdot 2 = 0, \\[4pt] &\lim_{x \to 1^+} (x - 1)\,f(x) = 0 \cdot (-1) = 0. \end{aligned}

\lim_{x \to 1} (x - 1)\,f(x) = 0

. A factor that approaches

0

wins over a jump, as long as the other factor has a finite limit from each side.

Composite functions

The second condition says the graph of

f

has no break at

L

. Then the limit passes inside

f

: find the limit of the inside function, and evaluate

f

there.

Example 5

Let

f(u) = \begin{cases} \sqrt{u}, & 0 \le u < 4 \\ u - 1, & u \ge 4 \end{cases}

and let

g(x) = 4 - (x - 1)^2

and

h(x) = 4 + (x - 1)^2

. Find

\lim_{x \to 1} f(g(x))

and

\lim_{x \to 1} f(h(x))

, and compare the first with

f(g(1))

f jumps at u = 4: from the left it approaches 2, and f(4) = 3.

g stays below the dashed line y = 4 and h stays above it, except at x = 1.

The rule above needs

f

to have no break at

4

, and

f

jumps there, so look at which side of

4

the inside function is on.

x

approaches

1

g(x)

approaches

4

, and

g(x) < 4

for every

x \ne 1

. So

f(g(x)) = \sqrt{g(x)}

there, and

\sqrt{u}

has no break at

4

, so the rule above applies to it:

\begin{aligned} &\lim_{x \to 1} f\bigl(g(x)\bigr) \\[4pt] &\qquad = \lim_{x \to 1} \sqrt{g(x)} = \sqrt{4} = 2. \end{aligned}

For

h

h(x) > 4

for every

x \ne 1

, so

f(h(x)) = h(x) - 1

there:

\begin{aligned} &\lim_{x \to 1} f\bigl(h(x)\bigr) \\[4pt] &\qquad = \lim_{x \to 1} \bigl(h(x) - 1\bigr) = 4 - 1 = 3. \end{aligned}

x = 1

itself,

f(g(1)) = f(4) = 3

, which differs from the limit

2

. Since

g(1) = 4

is also the inside limit, this is evaluating

f

at the inside limit, which the rule justifies only when

f

has no break there. For

h

it happens to give the right answer, because

f(4)

comes from the right-hand piece and

h

approaches

4

from the right. At a jump, that shortcut can't be trusted.

Example 6

Let

k(x) = x - 2

for

x < 1

and

k(x) = x + 2

for

x \ge 1

, and let

p(u) = u^2 - 2u

. Find

\lim_{x \to 1} p(k(x))

The inside function

k

has no limit at

1

, so split by sides. From the left,

k(x) = x - 2

approaches

-1

. From the right,

k(x) = x + 2

approaches

3

. The polynomial

p

has no break anywhere, so the rule above works one side at a time:

\begin{aligned} &\lim_{x \to 1^-} p\bigl(k(x)\bigr) = p(-1) = 1 + 2 = 3, \\[8pt] &\lim_{x \to 1^+} p\bigl(k(x)\bigr) = p(3) = 9 - 6 = 3. \end{aligned}

The one-sided limits agree because

p(-1) = p(3)

, so

\lim_{x \to 1} p(k(x)) = 3

, even though

k

has no limit at

1

. With

p(u) = u^2

instead, the sides would give

1

and

9

, and the limit wouldn't exist.