1.6 - Determining Limits Using Algebraic Manipulation

Method 1: Direct Substitution

The simplest way to find a limit is direct substitution: plug the

x

-value straight into the function. It works whenever plugging in gives you a defined result (not

\frac{0}{0}

or some other undefined form).

Example 1: polynomial function

Find

\lim_{x \to 2} (x^2 - 3x + 5)

\lim_{x \to 2} (x^2 - 3x + 5) = (2)^2 - 3(2) + 5 = 4 - 6 + 5 = 3

Answer:

3

1.6 - Determining Limits Using Algebraic Manipulation

x

approaches 2, the function smoothly approaches

y = 3

Example 2: constant function

Find

\lim_{x \to 5} 8

A constant function always gives the same output no matter the input:

\lim_{x \to 5} 8 = 8

For any constant

c

\lim_{x \to a} c = c

Method 2: Factor and Cancel

When direct substitution gives

\frac{0}{0}

(the indeterminate form), we have to manipulate the expression algebraically.

If substituting

x = a

into

\frac{f(x)}{g(x)}

gives

\frac{0}{0}

, that tells us: - Both numerator and denominator equal zero at

x = a

- So

(x - a)

is probably a common factor - We can factor and cancel to simplify

Example 3: factor out common terms

Find

\lim_{x \to 0} \frac{3x^2 - 7x}{x}

Step 1, try direct substitution:

\frac{3(0)^2 - 7(0)}{0} = \frac{0}{0} \quad \text{(indeterminate)}

Step 2, factor the numerator:

\lim_{x \to 0} \frac{3x^2 - 7x}{x} = \lim_{x \to 0} \frac{x(3x - 7)}{x}

Step 3, cancel the common factor:

= \lim_{x \to 0} (3x - 7)

Step 4, direct substitution:

= 3(0) - 7 = -7

Answer:

-7

The graph is a line with a hole at

x = 0

, but the limit still exists because both sides approach

-7

Example 4: factoring quadratics

Find

\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Step 1, check for the indeterminate form:

\frac{(3)^2 - 9}{3 - 3} = \frac{0}{0}

Step 2, factor using difference of squares:

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

Step 3, cancel

(x - 3)

= \lim_{x \to 3} (x + 3)

Step 4, substitute:

= 3 + 3 = 6

Answer:

6

Example 5: factoring with non-obvious factors

Find

\lim_{x \to -5} \frac{3x^2 + 16x + 5}{x + 5}

Step 1, substituting

x = -5

gives

\frac{0}{0}

Step 2, factor the numerator. Since the denominator is

(x + 5)

, we know

(x + 5)

has to be a factor of the numerator:

3x^2 + 16x + 5 = (x + 5)(3x + 1)

Check:

(x + 5)(3x + 1) = 3x^2 + x + 15x + 5 = 3x^2 + 16x + 5

✓

Step 3, simplify:

\lim_{x \to -5} \frac{(x + 5)(3x + 1)}{x + 5} = \lim_{x \to -5} (3x + 1)

Step 4, substitute:

= 3(-5) + 1 = -15 + 1 = -14

Answer:

-14

Method 3: Recognizing When Limits Don’t Exist

Not every limit exists. After factoring and simplifying, if you still can’t evaluate it, it probably doesn’t exist.

Example 6: vertical asymptote

Find

\lim_{x \to 4} \frac{x^2 + 2x + 8}{(x - 4)(x + 1)}

Step 1, try direct substitution:

\frac{(4)^2 + 2(4) + 8}{(4 - 4)(4 + 1)} = \frac{32}{0} \quad \text{(undefined)}

Step 2, try to factor

x^2 + 2x + 8

. We need two numbers that multiply to 8 and add to 2. There aren’t any real numbers that do that, so the numerator doesn’t factor nicely.

Step 3, since we can’t cancel the

(x - 4)

in the denominator, the expression can’t be simplified.

Answer: DNE

The left and right sides approach different infinities, so the limit doesn’t exist.

Special Trigonometric Limits

There are two fundamental trig limits you have to memorize.

\boxed{\lim_{x \to 0} \frac{\sin x}{x} = 1}

\boxed{\lim_{x \to 0} \frac{1 - \cos x}{x} = 0}

Important conditions: - These only work when

x \to 0

- The variable must match in numerator and denominator

Equivalent forms

These also work in reciprocal form:

\lim_{x \to 0} \frac{x}{\sin x} = 1 \qquad \lim_{x \to 0} \frac{\cos x - 1}{x} = 0

Memory tricks

For sine: the letter “s” in sine kind of looks like the number 1.

For cosine: the letter “c” in cosine kind of looks like the number 0.

(And remember, cosine limits involve

1 - \cos x

\cos x - 1

Working with Trig Limits

Example 7: sine limit with a coefficient

Find

\lim_{x \to 0} \frac{\sin 5x}{x}

We need the denominator to match what’s inside the sine.

Step 1, multiply by

\frac{5}{5}

\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} \frac{5 \cdot \sin 5x}{5x} = \lim_{x \to 0} 5 \cdot \frac{\sin 5x}{5x}

Step 2, apply the fundamental limit:

= 5 \cdot 1 = 5

Answer:

5

Quick pattern recognition

Shortcut: when you see

\lim_{x \to 0} \frac{\sin(kx)}{x}

, the answer is just

k

$\lim_{x \to 0} \frac{\sin 8x}{x} = 8$

$\lim_{x \to 0} \frac{\sin x}{3x} = \frac{1}{3}$

$\lim_{x \to 0} \frac{\sin 12x}{4x} = \frac{12}{4} = 3$

Example 8: quotient of two sine functions

Find

\lim_{x \to 0} \frac{\sin 6x}{\sin 4x}

The cleanest approach is to rewrite so each piece matches the fundamental form:

\lim_{x \to 0} \frac{\sin 6x}{\sin 4x} = \lim_{x \to 0} \frac{\sin 6x}{6x} \cdot \frac{6x}{4x} \cdot \frac{4x}{\sin 4x}

= 1 \cdot \frac{6}{4} \cdot 1 = \frac{3}{2}

Answer:

\frac{3}{2}

Example 9: cosine limit with factoring

Find

\lim_{x \to 0} \frac{\cos^2 x - 1}{x}

Step 1, factor using difference of squares:

\cos^2 x - 1 = (\cos x - 1)(\cos x + 1)

Step 2, rewrite:

\lim_{x \to 0} \frac{(\cos x - 1)(\cos x + 1)}{x} = \lim_{x \to 0} \frac{\cos x - 1}{x} \cdot (\cos x + 1)

Step 3, apply the fundamental cosine limit:

= 0 \cdot (1 + 1) = 0 \cdot 2 = 0

Answer:

0

Example 10: rewriting tangent

Find

\lim_{x \to 0} \frac{\tan 2x}{x}

Step 1, direct substitution gives

\frac{\tan 0}{0} = \frac{0}{0}

, so we need a different form.

Step 2, rewrite tangent in terms of sine and cosine:

\lim_{x \to 0} \frac{\tan 2x}{x} = \lim_{x \to 0} \frac{\sin 2x}{x} \cdot \frac{1}{\cos 2x}

Step 3, use the sine pattern (

\frac{\sin 2x}{x} \to 2

) and substitute into the cosine:

= 2 \cdot \frac{1}{\cos 0} = 2 \cdot 1 = 2

Answer:

2

Useful rewrites:

\tan x = \frac{\sin x}{\cos x}

\cot x = \frac{\cos x}{\sin x}

\sec x = \frac{1}{\cos x}

\csc x = \frac{1}{\sin x}

, and

\sin^2 x + \cos^2 x = 1

. When a trig limit doesn’t match the special forms, rewrite everything in sine and cosine first.

Practice Problems

Problem 1:

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

2(16) - 5(4) + 1 = 32 - 20 + 1 = 13

Problem 2:

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

Factor:

\frac{(x-2)(x+2)}{x-2} = x + 2 = 4

Problem 3:

\lim_{x \to 0} \frac{\sin 9x}{x} = 9

Problem 4:

\lim_{x \to 0} \frac{1 - \cos 3x}{x} = 0

(using the fundamental cosine limit)

Key Takeaways

Always try direct substitution first, it’s the fastest method

If you get $\frac{0}{0}$ , factor and cancel common terms

The denominator often hints at what factor to look for in the numerator

If you can’t simplify and still get $\frac{\text{nonzero}}{0}$ , the limit does not exist

Memorize the special trig limits:

$\lim_{x \to 0} \frac{\sin x}{x} = 1$
$\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$

Quick pattern: $\lim_{x \to 0} \frac{\sin kx}{x} = k$

Trig limits only work when $x \to 0$

Rewrite other trig functions in sine and cosine to reach the special forms

Difference of squares: $a^2 - b^2 = (a-b)(a+b)$

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