Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.) $\bigg\{\dfrac{e^n + e^{-n}}{e^{2n} - 1}\bigg\}$

$\displaystyle\lim_{n \to \infty} \bigg(\dfrac{e^n + e^{-n}}{e^{2n} - 1}\bigg) = \boxed{?}$

Neetesh Kumar | December 31, 2024

Calculus Homework Help

This is the solution to Math 1c
Assignment: 11.1 Question Number 34
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Step-by-step solution:

Step 1: Analyze the sequence:

The sequence is: $a_n = \dfrac{e^n + e^{-n}}{e^{2n} - 1}$

The numerator is: $e^n + e^{-n}$

The denominator is: $e^{2n} - 1$

Step 2: Divide numerator and denominator by $e^{2n}$:

To simplify, divide each term in the numerator and denominator by $e^{2n}$:

$a_n = \dfrac{\frac{e^n}{e^{2n}} + \frac{e^{-n}}{e^{2n}}}{\frac{e^{2n}}{e^{2n}} - \frac{1}{e^{2n}}}$

Simplify each term:

• $\frac{e^n}{e^{2n}} = e^{-n}$
• $\frac{e^{-n}}{e^{2n}} = e^{-3n}$
• $\frac{e^{2n}}{e^{2n}} = 1$
• $\frac{1}{e^{2n}} = e^{-2n}$

This gives:

$a_n = \dfrac{e^{-n} + e^{-3n}}{1 - e^{-2n}}$

Step 3: Evaluate the limit as $n \to \infty$:

As $n \to \infty$, the terms with negative exponents ($e^{-n}, e^{-2n}, e^{-3n}$) approach $0$:

• $e^{-n} \to 0$
• $e^{-2n} \to 0$
• $e^{-3n} \to 0$

Therefore:

• The numerator $e^{-n} + e^{-3n} \to 0$
• The denominator $1 - e^{-2n} \to 1$

Step 4: Conclude the limit:

Since the numerator approaches $0$ and the denominator approaches $1$, the limit is:

$\displaystyle\lim_{n \to \infty} a_n = \dfrac{0}{1} = 0$

Final Answer:

$\displaystyle\lim_{n \to \infty} \bigg(\dfrac{e^n + e^{-n}}{e^{2n} - 1}\bigg) = \boxed{0}$

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