Question

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_n=\frac{{75}^{n-1}}{{99}^n}+\frac{5^n\sin n}{8^n}$$

Short answer

Answer: The limit of the sequence as n goes to infinity is 0.

Step-by-step solution

Analyzing the first term

Let's analyze the first term, $$\frac{{75}^{n-1}}{{99}^n}$$. We can rewrite this term as: $$\frac{{75}^{n-1}}{{99}^n}=\frac{{75}^n}{75\cdot {99}^n}=\frac{{75}^n}{75\cdot ({99}^n)}$$ Now, by taking the limit as n goes to infinity, we get: $$\underset{n\to \mathrm{\infty }}{lim}\frac{{75}^n}{75\cdot {99}^n}$$

Applying the limit on the first term

To evaluate the limit for the first term, we can use the ratio property for sequence limits: $$\underset{n\to \mathrm{\infty }}{lim}\frac{{75}^n}{75\cdot {99}^n}=\frac{\underset{n\to \mathrm{\infty }}{lim}{75}^n}{\underset{n\to \mathrm{\infty }}{lim}75\cdot {99}^n}=\frac{\underset{n\to \mathrm{\infty }}{lim}(\frac{3}{4}{)}^n}{\underset{n\to \mathrm{\infty }}{lim}75\cdot (\frac{8}{8}{)}^n}$$ Since the limit of the ratio of sequences is the ratio of their limits, and $\frac{3}{4}<1$, we have $$\underset{n\to \mathrm{\infty }}{lim}\frac{{75}^n}{75\cdot {99}^n}=\frac{0}{75}=0$$

Analyzing the second term

Now, let's analyze the second term, $$\frac{5^n\sin n}{8^n}$$. We can rewrite this term as: $$\frac{5^n\sin n}{8^n}=(\frac{5}{8}{)}^n\cdot \sin (n)$$ Now, by taking the limit as n goes to infinity, we get: $$\underset{n\to \mathrm{\infty }}{lim}(\frac{5}{8}{)}^n\cdot \sin (n)$$

Applying the limit on the second term

To evaluate the limit for the second term, we can use the Squeeze theorem along with the fact that $\sin (n)$ is bounded between -1 and 1: $$-1\le \sin (n)\le 1$$ Then, we can multiply these inequalities by the positive term $(\frac{5}{8}{)}^n$: $$-(\frac{5}{8}{)}^n\le (\frac{5}{8}{)}^n\cdot \sin (n)\le (\frac{5}{8}{)}^n$$ Now taking the limit as n goes to infinity: $$\underset{n\to \mathrm{\infty }}{lim}-(\frac{5}{8}{)}^n\le \underset{n\to \mathrm{\infty }}{lim}(\frac{5}{8}{)}^n\cdot \sin (n)\le \underset{n\to \mathrm{\infty }}{lim}(\frac{5}{8}{)}^n$$ As $\frac{5}{8}<1$, the limits of the bounds are both 0: $$0\le \underset{n\to \mathrm{\infty }}{lim}(\frac{5}{8}{)}^n\cdot \sin (n)\le 0$$ By the Squeeze theorem, we conclude that the limit of the second term is also 0.

Calculating the limit of the sequence

As we have found the limits of both terms, we can now calculate the limit of the given sequence: $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}(\frac{{75}^{n-1}}{{99}^n}+\frac{5^n\sin n}{8^n})=\underset{n\to \mathrm{\infty }}{lim}\frac{{75}^{n-1}}{{99}^n}+\underset{n\to \mathrm{\infty }}{lim}\frac{5^n\sin n}{8^n}=0+0=0$$ As a result, the limit of the given sequence as n goes to infinity is 0.

Key concepts

Squeeze Theorem

The Squeeze Theorem, also known as the Sandwich Theorem, is an invaluable tool in calculus for determining the limits of sequences and functions. The beauty of the Squeeze Theorem lies in its ability to find the limit of a complicated sequence by 'squeezing' it between two simpler ones.

Imagine having a sequence that is difficult to analyze directly. If we can find two other sequences that bound our sequence from above and below, and if these bounding sequences converge to the same limit, then we can confidently say that our original sequence must converge to the same limit as well. This principle holds true as long as the 'squeezer' sequences remain consistently above and below our target sequence at all times, especially as we approach infinity.

How it's applied

Let's refer to the problem from the textbook solution, where we have a term involving $\frac{5^n\sin n}{8^n}$. Here, the Squeeze Theorem comes into play since $\sin n$ fluctuates between -1 and 1, which poses a challenge. However, since $(\frac{5}{8}{)}^n$ diminishes to zero as $n$ increases, and $\sin (n)$ is trapped between -1 and 1, we can multiply these bounds by $(\frac{5}{8}{)}^n$ to apply the Squeeze Theorem effectively. This results in proving that the limit of the complex term is also zero.

Limits of Sequences

Understanding the limits of sequences is foundational to mastering calculus and various other fields of mathematics. A limit of a sequence is, in essence, the value that the terms of the sequence approach as the index, usually denoted by $n$, goes to infinity.

In practical applications and theoretical constructs alike, calculating the limit of a sequence allows us to predict behavior at very high indices, which might represent time, quantity, or any other variable that is prone to incremental changes. A sequence can have a finite limit, an infinite limit, or it might oscillate indefinitely without settling on a particular value.

Practical example

In our exercise, we have two different sequences to consider. One is a geometric sequence with a ratio less than 1, $(\frac{3}{4}{)}^n$, which we know shrinks as $n$ becomes large, trending towards zero. The other sequence involves a trigonometric function, which complicates matters. Despite the complexity, we're able to reveal the limit by examining the behavior of each component and employing the Squeeze Theorem where appropriate.

Properties of Limits

Diving deeper into the world of sequences and calculus, properties of limits emerge as the governing rules that make solving limit problems a systematic process rather than a guessing game. These properties are based on an understanding that limits behave predictably under certain mathematical operations.

Some of these properties include the limit of a constant sequence being the constant itself, the limit of a sum being the sum of the limits, and the limit of a product being the product of the limits, provided that the limits exist. Moreover, we can divide the limits of sequences as long as the denominator doesn't approach zero.

Case in point

In the step-by-step solution, we see these properties in action. When separating the given sequence into two terms, the solution exploits these properties to find the limit of each term independently. The first term is a direct application of dividing limits, while the second term combines the limit of a product with the Squeeze Theorem. It is the clear understanding and application of these properties that allow us to conclude that the limit of the entire sequence is zero.