Problem 40 - Single Variable Calculus

Question

$$ \text { Let } g(x)=\frac{\cos x}{\sqrt{x}} . \text { Find } \lim _{x \rightarrow \infty} g(x) \text { . } $$

Step-by-Step Solution

Answer

The limit of g(x) as x approaches infinity is 0, expressed mathematically as: \[\lim_{x\rightarrow \infty} \frac{\cos x}{\sqrt{x}} = 0\]

Step 1: Analyze the behavior of \(\cos x\) as \(x \rightarrow \infty\)

As x approaches infinity, the value of the cosine function oscillates between -1 and 1. Thus, we cannot make any definitive conclusions about the behavior of \( \cos x\) as \(x \rightarrow \infty\).

Step 2: Analyze the behavior of \(\sqrt{x}\) as \(x \rightarrow \infty\)

As x approaches infinity, the value of \(\sqrt{x}\) also approaches infinity. This indicates that the denominator of our function will become larger and larger as x increases.

Step 3: Determine the limit of g(x) as \(x \rightarrow \infty\)

As we found out earlier, the cosine function oscillates between -1 and 1, while the denominator \(\sqrt{x}\) approaches infinity. Since the numerator is always between -1 to 1 and the denominator is increasing indefinitely, we can conclude that the limit of \(\frac{\cos x}{\sqrt{x}}\) as \(x \rightarrow \infty\) is 0. Mathematically, we express this as: \[\lim_{x\rightarrow \infty} g(x) = \lim_{x\rightarrow \infty} \frac{\cos x}{\sqrt{x}} = 0\] So, the limit of g(x) as x approaches infinity is 0.

Key Concepts

Cosine function

The cosine function, denoted as \( \cos x \), is a periodic trigonometric function. It is known for its wave-like behavior due to its constant oscillation between the values of -1 and 1. This oscillation occurs because cosine is defined based on the unit circle, where it represents the horizontal coordinate of a point moving around the circle. The frequency of this oscillation is consistent, with a complete cycle repeating every

• \(2\pi\): a full rotation around the circle translates to 360 degrees.

For any real number \( x \), the angle on the unit circle corresponding to \( \cos x \) remains between these bounds. Therefore, regardless of how large \( x \) becomes, \( \cos x \) will always fluctuate within this fixed range. This characteristic is crucial in evaluating limits involving infinity, as it provides a stable, bounded numerator when combined with unbounded functions in denominators.

Square root function

The square root function, expressed as \( \sqrt{x} \), operates by taking the non-negative square root of the variable \( x \). In mathematical terms, for a non-negative input \( x \), the square root function returns a value \( y \) such that \( y^2 = x \). At first glance, this function might seem simple, but its properties play a significant role when considering limits approaching infinity.

• As \( x \) increases without bound, \( \sqrt{x} \) also increases.
• However, the growth rate of \( \sqrt{x} \) is slower than that of \( x \) itself.

This characteristic of slower growth can profoundly affect expressions where it's in the denominator. As \( x \) tends towards infinity, the square root also tends toward infinity but at a moderated pace. This means in a function like \( \frac{1}{\sqrt{x}} \), numerical values decrease, approaching zero, which has implications when finding limits.

Behavior of functions at infinity

When analyzing functions as their inputs approach infinity, we are interested in understanding what the output does as the input grows larger and larger. For a function such as \( g(x) = \frac{\cos x}{\sqrt{x}} \), this process involves considering both the numerator and denominator separately:

• The numerator, \( \cos x \), remains bounded between -1 and 1 because it oscillates.
• The denominator, \( \sqrt{x} \), approaches infinity since it continuously increases with increasing \( x \).

In essence, the behavior of the denominator predominates because it becomes much larger as \( x \) increases, overpowering the bounded nature of the numerator. This relationship between a bounded numerator and an increasing denominator leads to an understanding that the overall function \( \frac{\cos x}{\sqrt{x}} \) diminishes towards zero. This is why, mathematically, the limit \( \lim_{x \rightarrow \infty} \frac{\cos x}{\sqrt{x}} = 0 \). Functions behaving this way at infinity can simplify otherwise complex limit evaluations, allowing the use of dominant behavior in denominators to draw conclusions.