1.5 - Determining Limits Using Algebraic Properties of Limits

Introduction

Just like regular numbers and variables have algebraic properties, so do limits. These properties let us break complicated limit problems into simpler pieces.

If you remember from algebra that

x + x = 2x

, limit properties work the same way:

\lim_{x \to c} f(x) + \lim_{x \to c} f(x) = 2 \cdot \lim_{x \to c} f(x)

Fundamental Limit Properties

Let

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

and

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

, where

L

and

M

are real numbers.

Sum

\lim_{x \to c} [f(x) + g(x)] = \lim_{x \to c} f(x) + \lim_{x \to c} g(x) = L + M

Difference

\lim_{x \to c} [f(x) - g(x)] = \lim_{x \to c} f(x) - \lim_{x \to c} g(x) = L - M

Constant multiple

(where

k

is any constant)

\lim_{x \to c} [k \cdot f(x)] = k \cdot \lim_{x \to c} f(x) = k \cdot L

Product

\lim_{x \to c} [f(x) \cdot g(x)] = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x) = L \cdot M

Quotient

\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)} = \frac{L}{M}, \quad \text{provided } M \neq 0

Power

\lim_{x \to c} [f(x)]^n = \left[\lim_{x \to c} f(x)\right]^n = L^n

Composite

(when

f

is continuous at

M

)

\lim_{x \to c} f(g(x)) = f\left(\lim_{x \to c} g(x)\right) = f(M)

For example,

\lim_{x \to 2} \sqrt{x^2 + 5} = \sqrt{\lim_{x \to 2} (x^2 + 5)} = \sqrt{9} = 3

. The square root is continuous at 9, so the limit can move inside.

Working with Tables of Values

When you’re given tables showing limits and function values, pay close attention to which one you’re pulling from.

Example 1: sum property

$x$	$\lim_{x \to 2} f(x)$	$\lim_{x \to 2} g(x)$
2	5	3

Find

\lim_{x \to 2} [f(x) + g(x)]

\lim_{x \to 2} [f(x) + g(x)] = \lim_{x \to 2} f(x) + \lim_{x \to 2} g(x) = 5 + 3 = 8

Example 2: inner functions with variables

	$\lim_{x \to -3}$	$\lim_{x \to 3}$
$f(x)$	4	7
$g(x)$	-2	5

Find

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

Step 1: figure out what the inner function does. As

x \to 3

, we have

-x \to -3

Step 2: so we need

\lim_{x \to -3} f(x)

for the first piece and

\lim_{x \to 3} g(x)

for the second:

\lim_{x \to 3} [f(-x) + g(x)] = \lim_{x \to -3} f(x) + \lim_{x \to 3} g(x) = 4 + 5 = 9

Composition of Functions

With composite functions like

f(g(x))

, work from the inside out.

Example 3: nested functions

$x$	2	6
$\lim_{x \to 2} f(x)$	6	-
$\lim_{x \to 6} f(x)$	-	10

Find

\lim_{x \to 2} f(f(x))

Step 1: evaluate the inner limit first.

\lim_{x \to 2} f(x) = 6

Step 2: the inner function approaches 6, so now we need

\lim_{x \to 6} f(x)

. In other words, what does

f(x)

approach as

x

approaches 6?

Step 3: from the table:

\lim_{x \to 6} f(x) = 10

\lim_{x \to 2} f(f(x)) = 10

Think of it this way: we plug

x = 2

into the inner

f

and get an output approaching 6. Then we use 6 as the new input and get an output approaching 10.

Heads up: on graph problems, check how the inner function approaches 6. If its outputs approach 6 from above (like 6.1, 6.01, ...), use the right-hand limit

\lim_{x \to 6^+} f(x)

. If they approach from below, use the left-hand limit. And if the inner function is flat and actually equals 6 near

x = 2

, the outer function gets exactly 6 as its input, so use the value

f(6)

. These only give different answers when the outer function isn’t continuous at 6.

Function Values vs. Limits

There’s a real difference between

f(a)

(the actual function value) and

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

(the limit). Don’t mix them up.

Example 4: mixed function values and limits

$x$	$f(x)$	$g(x)$	$h(x)$	$\lim_{x \to 3} f(x)$	$\lim_{x \to 3} g(x)$	$\lim_{x \to 3} h(x)$
3	2	5	8	10	4	6

Find

\lim_{x \to 3} [h(x) \cdot (f(x) + 2g(x))] - h(3)

Step 1: identify the limit portion (everything before the subtraction):

\lim_{x \to 3} [h(x) \cdot (f(x) + 2g(x))]

Step 2: break it apart with the properties:

= \lim_{x \to 3} h(x) \cdot \lim_{x \to 3} [f(x) + 2g(x)]

= \lim_{x \to 3} h(x) \cdot \left[\lim_{x \to 3} f(x) + 2 \cdot \lim_{x \to 3} g(x)\right]

Step 3: substitute values from the limit columns:

= 6 \cdot [10 + 2(4)] = 6 \cdot [10 + 8] = 6 \cdot 18 = 108

Step 4: subtract the function value

h(3)

. From the table,

h(3) = 8

(function value column, not the limit column):

108 - 8 = 100

Answer:

100

Piecewise Functions and Limits

Piecewise functions need extra care because the formula changes at certain points.

f(x) = \begin{cases} \text{Formula 1} & \text{if } x < a \\ \text{Formula 2} & \text{if } x \geq a \end{cases}

For $\lim_{x \to a^-} f(x)$ , use Formula 1 (approaching from the left)

For $\lim_{x \to a^+} f(x)$ , use Formula 2 (approaching from the right)

$\lim_{x \to a} f(x)$ exists only if both one-sided limits are equal

Example 5: piecewise limits

g(x) = \begin{cases} \sqrt{8 - x} & \text{if } x < -3 \\ \dfrac{x + 2}{4 - x^2} & \text{if } x \geq -3 \end{cases}

Find (a)

\lim_{x \to -3^-} g(x)

, (b)

\lim_{x \to -3^+} g(x)

, (c)

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

Part (a): left-hand limit

Approaching from the left means

x < -3

, so use the first piece:

\lim_{x \to -3^-} g(x) = \sqrt{8 - (-3)} = \sqrt{8 + 3} = \sqrt{11}

On a calculator,

\sqrt{11} \approx 3.317

Answer:

\sqrt{11}

, or

3.317

Part (b): right-hand limit

Approaching from the right means

x > -3

(just barely), so use the second piece:

\lim_{x \to -3^+} g(x) = \frac{-3 + 2}{4 - (-3)^2} = \frac{-1}{4 - 9} = \frac{-1}{-5} = \frac{1}{5}

Answer:

\frac{1}{5}

, or

0.2

Part (c): two-sided limit

For this to exist, both one-sided limits must match.

Left-hand limit: $\sqrt{11} \approx 3.317$

Right-hand limit: $\frac{1}{5} = 0.2$

Since

\sqrt{11} \neq \frac{1}{5}

, they don’t match.

Answer: DNE

Practice Problems

Problem 1

$x$	$\lim_{x \to 4} f(x)$	$\lim_{x \to 4} g(x)$
4	12	-3

Find

\lim_{x \to 4} [2f(x) - 5g(x)]

\lim_{x \to 4} [2f(x) - 5g(x)] = 2(12) - 5(-3) = 24 + 15 = 39

Problem 2

h(x) = \begin{cases} 2x + 5 & \text{if } x < 1 \\ x^2 + 4 & \text{if } x \geq 1 \end{cases}

Find

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

Left-hand limit:

\lim_{x \to 1^-} h(x) = 2(1) + 5 = 7

Right-hand limit:

\lim_{x \to 1^+} h(x) = (1)^2 + 4 = 5

Since

7 \neq 5

, the limit does not exist.

Problem 3

$x$	$f(x)$	$\lim_{x \to 5} f(x)$
5	3	8

And

g(x) = \ln(x)

. Find

\lim_{x \to 5} [g(f(x))]

Inner limit:

\lim_{x \to 5} f(x) = 8

\lim_{x \to 5} [g(f(x))] = g(8) = \ln(8) \approx 2.079

Problem 4

p(x) = \begin{cases} \ln(x) & \text{if } x < e \\ 5 & \text{if } x \geq e \end{cases}

$\lim_{x \to e^-} p(x) = \ln(e) = 1$

$\lim_{x \to e^+} p(x) = 5$

$\lim_{x \to e} p(x) = \text{DNE}$ , since $1 \neq 5$

Key Takeaways

Limit properties mirror algebraic properties (sum, difference, product, quotient)

Always work inside out with composite functions

Distinguish between $f(a)$ (function value) and $\lim_{x \to a} f(x)$ (limit)

For piecewise functions:

Left-hand limits use the piece where $x <$ boundary
Right-hand limits use the piece where $x >$ or $x \geq$ boundary

Two-sided limits exist only when both one-sided limits are equal

With tables, carefully check whether you need function values or limits

The quotient property requires the denominator’s limit to be nonzero

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