Day 27: The Definition of a Limit and Continuity & Relative Growth Rates & Horizontal Asymptotes (HA) & Intermediate Value Theorem (IVT)

The Definition of a Limit and Continuity

Understanding that a limit exists only if the behavior from the left matches the behavior from the right, and continuity requires the limit to equal the function value.

Core Theorem

A function

f(x)

is continuous at

x=c

if and only if: 1.

f(c)

is defined. 2.

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

exists (meaning

\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)

). 3.

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

.

Step-by-Step SOP

1
Check Left and Right
Always check $\lim_{x \to c^-}$ and $\lim_{x \to c^+}$ separately for piecewise functions.
2
Verify the Point
Ensure $f(c)$ exists and matches the limit for continuity.

Practice Exercises

Example 01Easy

Let

f(x) = \begin{cases} x^2 + k & x < 2 \\ 3x - 1 & x \geq 2 \end{cases}

. Find the value of

k

that makes

f(x)

continuous at

x=2

.

NEED A HINT?

Set the left-hand limit equal to the right-hand limit at

x=2

.

SHOW DETAILED EXPLANATION

Step 1: Evaluate the Right-Hand Limit

As

x \to 2^+

, use the expression

3x - 1

. So,

\lim_{x \to 2^+} f(x) = 3(2) - 1 = 5

.

Step 2: Evaluate the Left-Hand Limit

As

x \to 2^-

, use the expression

x^2 + k

. So,

\lim_{x \to 2^-} f(x) = 2^2 + k = 4 + k

.

Step 3: Solve for k

For the limit to exist and the function to be continuous, Left = Right:

4 + k = 5 \implies k = 1

.

Example 02Medium

Determine if

\lim_{x \to 3} \frac{|x-3|}{x-3}

exists.

NEED A HINT?

Absolute value functions are 'V-shaped' and often have different slopes on either side of the vertex.

SHOW DETAILED EXPLANATION

Step 1: Test the Right Side

For

x > 3

,

|x-3| = x-3

. Therefore,

\lim_{x \to 3^+} \frac{x-3}{x-3} = 1

.

Step 2: Test the Left Side

For

x < 3

,

|x-3| = -(x-3)

. Therefore,

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

.

Step 3: Compare and Conclude

Since the left limit (-1) does not equal the right limit (1), the limit does not exist (DNE).

Example 03Easy

If

\lim_{x \to a} f(x) = 7

and

f(a) = undefined

, what type of discontinuity is at

x=a

?

NEED A HINT?

If the limit exists but the point is missing or misplaced, what do we call that?

SHOW DETAILED EXPLANATION

Conclusion

Because the limit exists (the two sides meet), but the function value is not there, it is a **Removable Discontinuity** (a hole).

Common Pitfalls

⚠
The 'Value' ConfusionA limit tells you where the function is *heading*, not where it *is*. A function can have a limit at a point where it is undefined.
⚠
Assuming ContinuityNever assume $f(c) = \lim_{x \to c} f(x)$ unless the problem explicitly states the function is continuous.

Relative Growth Rates

A shortcut method to evaluate limits as $x$ approaches infinity by comparing how fast functions grow.

Core Theorem

The Hierarchy of Growth: For

x \to \infty

, the relative speeds are:

e^x > x^p > \ln x

. Formally,

\lim_{x \to \infty} \frac{\text{Slower}}{\text{Faster}} = 0

and

\lim_{x \to \infty} \frac{\text{Faster}}{\text{Slower}} = \infty

.

Step-by-Step SOP

1
Rank the Functions
Label terms as Log, Power, or Exponential.
2
Identify the Winner
If the 'winner' is on the bottom, the limit is 0. If on top, it is $\infty$ .

Practice Exercises

Example 01Medium

Evaluate

\lim_{x \to \infty} \frac{x^{100}}{1.01^x}

.

NEED A HINT?

Compare a power function (

x^p

) to an exponential function (

a^x

).

SHOW DETAILED EXPLANATION

Step 1: Identify the Types

The numerator is a polynomial (power function) and the denominator is an exponential function.

Step 2: Apply Hierarchy

Even though

x^{100}

looks huge,

1.01^x

(exponential) will eventually grow much faster as

x \to \infty

.

Step 3: Final Answer

Since the denominator is 'faster', the ratio approaches 0.

Example 02Easy

Evaluate

\lim_{x \to \infty} \frac{\ln(x^{500})}{x}

.

NEED A HINT?

Use log properties first, then compare growth rates.

SHOW DETAILED EXPLANATION

Step 1: Simplify

\ln(x^{500}) = 500 \ln x

.

Step 2: Compare

We are comparing

500 \ln x

(Logarithmic) vs.

x

(Linear/Polynomial).

Step 3: Result

Polynomials grow faster than logs. The limit is 0.

Example 03Hard

Evaluate

\lim_{x \to \infty} \frac{e^x + x^2}{5e^x - \ln x}

.

NEED A HINT?

Identify the 'dominant' term in both the numerator and denominator.

SHOW DETAILED EXPLANATION

Step 1: Find Dominant Terms

In the numerator,

e^x

dominates

x^2

. In the denominator,

5e^x

dominates

\ln x

.

Step 2: Reduce

The limit behaves like

\lim_{x \to \infty} \frac{e^x}{5e^x}

.

Step 3: Simplify

The

e^x

terms cancel, leaving

1/5

.

Common Pitfalls

⚠
The Large Constant TrapDo not be fooled by large exponents like $x^{1000}$ . Any exponential $a^x$ (where $a > 1$ ) will eventually surpass it.
⚠
Approaching ZeroGrowth rate hierarchy only applies as $x \to \infty$ . It does not apply as $x \to 0$ .

Horizontal Asymptotes (HA)

Using the degrees of rational functions to determine the behavior of a function at the far ends of the x-axis.

Core Theorem

For

f(x) = \frac{N(x)}{D(x)}

: 1. If deg(N) < deg(D), HA is

y=0

. 2. If deg(N) = deg(D), HA is

y = \frac{\text{leading coef N}}{\text{leading coef D}}

. 3. If deg(N) > deg(D), no HA (limit is

\pm\infty

).

Step-by-Step SOP

1
Degree Check
Quickly identify the highest power in the top and bottom.
2
Coefficient Ratio
If degrees match, divide the leading coefficients.

Practice Exercises

Example 01Easy

Find the horizontal asymptote of

f(x) = \frac{4x^2 - 5x + 1}{2x^2 + 9}

.

NEED A HINT?

Check the highest power of

x

in the numerator and denominator.

SHOW DETAILED EXPLANATION

Step 1: Compare Degrees

Numerator degree is 2. Denominator degree is 2. They are equal.

Step 2: Take the Ratio

The leading coefficients are 4 and 2. Ratio =

4/2 = 2

.

Step 3: State the HA

The HA is the line

y=2

.

Example 02Hard

Evaluate

\lim_{x \to \infty} \frac{\sqrt{9x^2 + 1}}{2x - 5}

.

NEED A HINT?

Remember that

\sqrt{x^2}

acts like

|x|

. Consider the signs for

\infty

vs

-\infty

.

SHOW DETAILED EXPLANATION

Step 1: Analyze the Degree

The numerator effectively has degree 1 (since

\sqrt{x^2} = x

). The denominator has degree 1.

Step 2: Leading Coefficients

Numerator coefficient:

\sqrt{9} = 3

. Denominator coefficient: 2.

Step 3: Solve

The limit is

3/2

.

Example 03Medium

Evaluate

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

.

NEED A HINT?

The function is 'top-heavy'. Check the sign of the infinity.

SHOW DETAILED EXPLANATION

Step 1: Compare Degrees

Degree 3 (Top) > Degree 2 (Bottom). The limit will be

\infty

or

-\infty

.

Step 2: Check Sign

Numerator:

(-\infty)^3

is negative. Denominator:

(-\infty)^2

is positive. Negative/Positive = Negative.

Step 3: Result

The limit is

-\infty

.

Common Pitfalls

⚠
The Square Root TrapWhen $x \to -\infty$ , $\sqrt{x^2}$ results in a positive value, but the $x$ in the denominator stays negative. This often creates a negative HA.
⚠
Ignoring Small TermsOnly the highest power matters at infinity. Don't waste time calculating limits for lower-degree terms.

Intermediate Value Theorem (IVT)

A 'existence' theorem used to prove that a function must pass through a specific value.

Core Theorem

If

f

is continuous on

[a, b]

and

k

is any number between

f(a)

and

f(b)

, then there exists at least one number

c

in

(a, b)

such that

f(c) = k

.

Step-by-Step SOP

1
State Continuity
Confirm the function has no holes or jumps in the interval.
2
Test Endpoints
Plug in $a$ and $b$ to find the range of y-values.
3
Conclude
Show that the target value $k$ is between $f(a)$ and $f(b)$ .

Practice Exercises

Example 01Easy

Show that

f(x) = x^3 + x - 1

has a zero on the interval

[0, 1]

.

NEED A HINT?

Find

f(0)

and

f(1)

and check if 0 lies between them.

SHOW DETAILED EXPLANATION

Step 1: Check Continuity

f(x)

is a polynomial, so it is continuous everywhere.

Step 2: Evaluate Endpoints

f(0) = 0^3 + 0 - 1 = -1

.

f(1) = 1^3 + 1 - 1 = 1

.

Step 3: Apply IVT

Since

f(0) < 0 < f(1)

, by IVT, there must be a

c \in (0, 1)

such that

f(c) = 0

.

Example 02Easy

A continuous function

g

has values

g(2) = 10

and

g(5) = 20

. Is there a value

x=c

such that

g(c) = 15

?

NEED A HINT?

Is 15 between 10 and 20?

SHOW DETAILED EXPLANATION

Analysis

Since

g

is continuous and

10 < 15 < 20

, the IVT guarantees at least one

c

in

(2, 5)

where

g(c) = 15

.

Example 03Medium

Why can't IVT be used for

f(x) = 1/x

on

[-1, 1]

to prove

f(c) = 0

?

NEED A HINT?

Check the requirements of the theorem.

SHOW DETAILED EXPLANATION

Conclusion

f(x) = 1/x

is not continuous at

x=0

, which is inside the interval

[-1, 1]

. The continuity requirement is violated.

Common Pitfalls

⚠
Forgetting ContinuityYou MUST state that the function is continuous before using IVT on an exam, or you will lose points.
⚠
The x vs y TrapIVT guarantees a 'c' value (an x-value), but you use the 'y-values' (endpoints) to prove it.

Check Left and Right

Verify the Point

Practice Exercises

Step 1: Evaluate the Right-Hand Limit

Step 2: Evaluate the Left-Hand Limit

Step 3: Solve for k

Step 1: Test the Right Side

Step 2: Test the Left Side

Step 3: Compare and Conclude

Conclusion

Rank the Functions

Identify the Winner

Practice Exercises

Step 1: Identify the Types

Step 2: Apply Hierarchy

Step 3: Final Answer

Step 1: Simplify

Step 2: Compare

Step 3: Result

Step 1: Find Dominant Terms

Step 2: Reduce

Step 3: Simplify

Degree Check

Coefficient Ratio

Practice Exercises

Step 1: Compare Degrees

Step 2: Take the Ratio

Step 3: State the HA

Step 1: Analyze the Degree

Step 2: Leading Coefficients

Step 3: Solve

Step 1: Compare Degrees

Step 2: Check Sign

Step 3: Result

State Continuity

Test Endpoints

Conclude

Practice Exercises

Step 1: Check Continuity

Step 2: Evaluate Endpoints

Step 3: Apply IVT

Analysis

Conclusion

Gary Chang