Question

Evaluate the limit of the following sequences. $$a_{n}=\cos \left(0.99^{n}\right)+\frac{7^{n}+9^{n}}{63^{n}}$$

Short answer

Answer: The limit of the given sequence as n approaches infinity is 1.

Step-by-step solution

Analyze the cosine function

The first part of the sequence is: $$\cos \left(0.99^n\right)$$ As n approaches infinity, the term \(0.99^n\) will approach 0. Thus, the cosine function also approaches: $$\cos(0) = 1$$

Analyze the fraction with exponentials

The second part of the sequence is: $$\frac{7^n + 9^n}{63^n}$$ To analyze this term, we can rewrite it as: $$\frac{7^n}{63^n} + \frac{9^n}{63^n}$$ Now, we can simplify each term by dividing the exponents: $$\frac{1}{56^n}+\frac{1}{7^n}$$ As n approaches infinity, both of these terms will approach 0, since the denominator grows infinitely large.

Combine the results

Now, we can combine the results from Steps 1 and 2. $$\lim_{n\to\infty} a_n = \cos \left(0.99^n\right) + \frac{7^n + 9^n}{63^n}$$ $$= 1 + 0 + 0 = 1$$ Therefore, the limit of the given sequence is 1 as n approaches infinity.

Key concepts

Cosine Function

The cosine function is an essential concept in trigonometry. It is a periodic function, which means it repeats its values in regular intervals. The cosine of an angle gives the horizontal component of the angle's vector on the unit circle. For a given angle \(\theta\), the cosine can be expressed as \(\cos(\theta)\). This function oscillates between -1 and 1.

In the context of limits, the cosine function becomes particularly useful when dealing with very small angles or numbers approaching zero. For instance, in the sequence provided, \(0.99^n\) becomes tiny as \(n\) grows larger. The cosine of a number that approaches zero also moves towards \(\cos(0) = 1\). This is because for small values, the cosine function flattens into a nearly horizontal line at its \(y=1\) maximum. This characteristic is crucial for evaluating the sequence's limit as part of understanding how oscillations diminish with specific sequences.

Exponential Decay

Exponential decay refers to the process where values decrease rapidly at first and then slow down in rate over time. It is represented by functions where the variable appears as an exponent with a base less than one, such as \(\left(\frac{1}{b}\right)^n\), where \(b > 1\) and \(n\) increases.

In our sequence, terms like \(\frac{7^n}{63^n}\) and \(\frac{9^n}{63^n}\) naturally illustrate exponential decay. As \(n\) becomes very large, the numerator's growth is outpaced by the denominator's size, causing each fraction to approach zero. This occurs because the bases 7 and 9, being much smaller than 63, decay swiftly relative to the base 63. Understanding this decay is pivotal for grasping why these terms vanish at infinity, crucially lowering the sequence's second part toward zero.

Infinite Limits

Infinite limits occur when the value of a sequence or function approaches a specific number as its variable heads towards infinity. It is a fundamental idea in calculus that allows us to comprehend long-term behavior.

When evaluating limits like \(\lim_{n\to\infty} a_n\), we consider how the sequence behaves as \(n\) grows endlessly. In our exercise, we look at how two parts of the sequence change: the cosine component \(\cos(0.99^n)\) approximates to 1, while the exponential fractions \(\frac{1}{56^n}+\frac{1}{7^n}\) both trend towards zero. These behaviors clarify the sequence's destination - in this case, merging into the single value of 1 as \(n\) approaches infinity. This convergence is a perfect example of why studying infinite limits deepens our understanding of mathematical models and functions.