Sequence Limits Using Functions

Example 2: Finding the Limit of the Sequence 1/(n+1)

Sequence Representation

Consider the sequence given by 1/2, 1/3, 1/4, … This sequence can be represented by the function f(n) = 1/(n+1).

Finding the Limit Using the Function

To find the limit of this sequence, we can analyze the corresponding function as n approaches infinity.

• Define the function f(x) = 1/(x+1). This function represents the pattern of our sequence.
• Take the limit of the function as x → ∞: lim(x → ∞) 1/(x+1) = 0. As x gets larger and larger, the value of 1/(x+1) gets closer and closer to 0.

Therefore, the limit of the sequence 1/2, 1/3, 1/4, … is 0.

Connection to the Sequence

The limit of the sequence is the same as the limit of the function f(x) = 1/(x+1) as x approaches infinity. This example shows how we can use functions to understand the behavior of sequences and find their limits.

Example 3: Finding the Limit of the Sequence (3n + 2)/(n + 1)

Sequence Representation

Consider the sequence given by 5/2, 8/3, 11/4, … This sequence can be represented by the function f(n) = (3n + 2)/(n + 1).

Finding the Limit Using the Function

To find the limit of this sequence, we can analyze the corresponding function as n approaches infinity.

• Define the function f(x) = (3x + 2)/(x + 1). This function represents the pattern of our sequence.
• Multiply both the numerator and denominator by 1/x:

f(x) = (1/x)/(1/x) * (3x + 2)/(x + 1) – Multiplying by (1/x)/(1/x) which is 1

f(x) = (3x/x + 2/x) / (x/x + 1/x) – Distributing 1/x in both numerator and denominator

f(x) = (3 + 2/x) / (1 + 1/x) – Simplifying 3x/x to 3 and x/x to 1

• Take the limit of the function as x → ∞: lim(x → ∞) (3 + 2/x) / (1 + 1/x) = 3. As x gets larger and larger, the value of (3 + 2/x) / (1 + 1/x) gets closer and closer to 3.

Therefore, the limit of the sequence 5/2, 8/3, 11/4, … is 3.

Connection to the Sequence

The limit of the sequence is the same as the limit of the function f(x) = (3x + 2)/(x + 1) as x approaches infinity. This example shows how we can use functions to understand the behavior of sequences and find their limits, even when the limit is not 0.

Example: Finding the Limit of the Sequence ln(n)/n as n → ∞

Sequence Representation

Consider the sequence given by ln(1)/1, ln(2)/2, ln(3)/3, … This sequence can be represented by the function f(n) = ln(n)/n.

Finding the Limit Using the Function

To find the limit of this sequence, we can analyze the corresponding function as n approaches infinity.

• Define the function f(x) = ln(x)/x. This function represents the pattern of our sequence.
• Consider the limit of the function as x → ∞:

lim(x → ∞) ln(x)/x

This limit is an indeterminate form (0/0), so we can use L’Hôpital’s Rule:

Take the derivative of the numerator and denominator:

d/dx [ln(x)] = 1/x

d/dx [x] = 1

So the limit becomes:

lim(x → ∞) (1/x) / 1

Since the denominator is 1, we can simplify this expression:

(1/x) / 1 = 1/x

Now, as x approaches infinity, the value of 1/x gets closer and closer to 0:

1/∞ = 0 (not official math, but helps to understand the concept)

So the limit is:

lim(x → ∞) 1/x = 0

Therefore, the limit of the sequence ln(1)/1, ln(2)/2, ln(3)/3, … is 0.

Connection to the Sequence

The limit of the sequence is the same as the limit of the function f(x) = ln(x)/x as x approaches infinity. This example shows how we can use functions and calculus techniques like L’Hôpital’s Rule to understand the behavior of sequences and find their limits.

Example: Finding the Limit of the Sequence (-1)ⁿ/n as n → ∞

Sequence Representation

Consider the sequence given by (-1)/1, 1/2, (-1)/3, 1/4, … This sequence can be represented by the function f(n) = (-1)ⁿ/n.

Finding the Limit Using the Function

To find the limit of this sequence, we can analyze the corresponding function as n approaches infinity.

• Define the function f(x) = (-1)ˣ/x. This function represents the pattern of our sequence.
• Consider the limit of the absolute value of the function as x → ∞:

|f(x)| = |(-1)ˣ/x|

Since the absolute value of (-1)ˣ is always 1, we can simplify:

|f(x)| = 1/x

Now, as x approaches infinity, the value of 1/x gets closer and closer to 0:

lim(x → ∞) 1/x = 0

• Since the absolute value of the function approaches 0, the function itself must also approach 0:

lim(x → ∞) (-1)ˣ/x = 0

Therefore, the limit of the sequence (-1)/1, 1/2, (-1)/3, 1/4, … is 0.

Connection to the Sequence

The limit of the sequence is the same as the limit of the function f(x) = (-1)ˣ/x as x approaches infinity. This example shows how we can use the concept of absolute value to understand the behavior of sequences and find their limits.

Continuity and Limits of Sequences

If we have a sequence \(a_n\) and its limit as \(n\) approaches infinity is \(L\), i.e.,

\(\lim_{n \to \infty} a_n = L\)

and if there is a function \(f\) that is continuous at \(L\), then the limit of the sequence formed by applying the function \(f\) to each term of the sequence \(a_n\) will be \(f(L)\):

\(\lim_{n \to \infty} f(a_n) = f(L)\)

This property connects the concept of continuity of functions with the limits of sequences. It allows us to understand how continuous functions behave when applied to sequences that converge to a specific value.

Example: Finding the Limit of ln(1/n) as n Approaches Infinity

We want to find the limit of the sequence given by aₙ = ln(1/n) as n approaches infinity.

• Start with the expression: lim(n → ∞) ln(1/n)
• Since the natural logarithm is continuous, we can move the limit inside: ln(lim(n → ∞) 1/n)
• Find the limit of the sequence inside the logarithm: lim(n → ∞) 1/n = 0
• Substitute the limit back into the expression: ln(0) = -∞

Therefore, the limit of the sequence aₙ = ln(1/n) as n approaches infinity is -∞.

Summary: lim(n → ∞) ln(1/n) = ln(lim(n → ∞) 1/n) = ln(0) = -∞

Example: Finding the Limit of sin(1/n) as n Approaches Infinity

We want to find the limit of the sequence given by aₙ = sin(1/n) as n approaches infinity.

• Start with the expression: lim(n → ∞) sin(1/n)
• Since the sine function is continuous, we can move the limit inside: sin(lim(n → ∞) 1/n)
• Find the limit of the sequence inside the sine function: lim(n → ∞) 1/n = 0
• Substitute the limit back into the expression: sin(0) = 0

Therefore, the limit of the sequence aₙ = sin(1/n) as n approaches infinity is 0.

Summary: lim(n → ∞) sin(1/n) = sin(lim(n → ∞) 1/n) = sin(0) = 0

Example: Finding the Limit of 1/n! as n Approaches Infinity

We want to find the limit of the sequence given by aₙ = 1/n! as n approaches infinity.

• Start with the expression: lim(n → ∞) 1/n!
• As n gets larger, the factorial n! grows much faster than any polynomial function of n. Therefore, the value of 1/n! gets closer and closer to 0.
• Since the sequence is decreasing and bounded below by 0, the limit is 0.

Therefore, the limit of the sequence aₙ = 1/n! as n approaches infinity is 0.

Summary: lim(n → ∞) 1/n! = 0

Convergence of the Sequence \( a_n – \frac{n!}{n^n} \)

Sequence Representation: \( a_n – \frac{1 \cdot 2 \cdot 3 \cdot \ldots \cdot n}{n \cdot n \cdot n \cdot \ldots \cdot n} \)

Observations: Numerator and denominator approach infinity, expression in parentheses is at most 1, terms are decreasing.

Inequality: \( 0 \leq a_n < 1 \)

Upper Bound Limit: \( \lim_{n \to \infty} 1 = 1 \)

Lower Bound Limit: \( \lim_{n \to \infty} 0 = 0 \)

Conclusion: \( \lim_{n \to \infty} a_n – \frac{n!}{n^n} = 0 \) by the Squeeze Theorem.