Problem 112 - Algebra and Trigonometry Real Mathematics, Real People

Question

Fill in the blanks. (Note: \(x \rightarrow c^{+}\) indicates that \(x\) approaches \(c\) from the right, and \(x \rightarrow c^{-}\) indicates that \(x\) approaches \(c\) from the left.) $$\text { As } x \rightarrow 0^{+}, \cos x \rightarrow$$ ______ $$\text { and } \sec x \rightarrow$$ ______

Step-by-Step Solution

Answer

As \(x \rightarrow 0^{+}\), \(\cos x \rightarrow 1 \) and \(\sec x \rightarrow 1\)

Step 1: Understanding the behavior of \(\cos x\) as \(x \rightarrow 0^{+}\)

As \(x\) gets closer and closer to 0 from the right, we look at the value of the cosine function \(\cos x\). Cosine of any angle lies between -1 and 1, and \(\cos(0) = 1\). So as \(x\) approaches 0 from the right, \(\cos x\) approaches 1.

Step 2: Filling the first blank

Based on our observations from Step 1, we can fill the first blank. As \(x \rightarrow 0^{+}\), \(\cos x \rightarrow 1\)

Step 3: Understanding the behaviour of \(\sec x\) as \(x \rightarrow 0^{+}\)

\(\sec x\) is reciprocal of \(\cos x\). Thus, as \(x \rightarrow 0^{+}\), \(\sec x = \frac{1}{\cos x}\) will approach the reciprocal of 1, which is 1 itself.

Step 4: Filling the second blank

Based on our observations from Step 3, we can fill the second blank. As \(x \rightarrow 0^{+}\), \(\sec x \rightarrow 1\)

Key Concepts

Cosine Function

The cosine function, denoted as \(\cos x\), is a fundamental trigonometric function that describes the ratio of the adjacent side to the hypotenuse in a right-angled triangle. When we look at the unit circle, which is a circle with a radius of one, the cosine of an angle \(x\) represents the x-coordinate of the point where the terminal side of the angle intersects the circle.

The cosine function is periodic with a period of \(2\pi\) and oscillates between -1 and 1. That means the output of \(\cos x\) will always be within this range, no matter what \(x\) value you have. This behavior is important when considering limits because it tells us that as \(x\) gets infinitesimally close to any given value, the cosine function's value will be predictable within this range.

When considering limits that approach 0 from the right, denoted as \(x \rightarrow 0^{+}\), the cosine function tends to the value it would have exactly at 0. Since \(\cos(0) = 1\), as \(x\) gets closer to 0 from the positive side, \(\cos x\) also gets closer to 1.

Secant Function

The secant function, represented as \(\sec x\), is the reciprocal of the cosine function, which means \(\sec x = \frac{1}{\cos x}\). Unlike the cosine, which has a range between -1 and 1, the range of the secant function is \( (-\infty, -1] \cup [1, \infty) \). This means that the secant function does not have any values between -1 and 1, and it approaches infinity as the cosine function approaches 0.

Since the secant function is defined as the reciprocal of the cosine, when \(\cos x\) is close to 1, \(\sec x\) will also be close to 1. Therefore, as we consider the limit of \(\sec x\) as \(x\) approaches 0 from the right, just like with the cosine function, \(\sec x\) approaches 1 too because \(\sec 0 = 1\). It's crucial to emphasize that the secant function will become undefined when \(\cos x = 0\), which happens at odd multiples of \(\frac{\pi}{2}\).

Limits Approaching from the Right

In calculus, evaluating the limit of a function as the variable approaches a particular value 'from the right' involves looking at values of the variable that are increasingly close to the given value, but strictly greater than it. This is denoted as \(x \rightarrow c^{+}\).

Approaching a limit from the right is crucial when the function has different behavior on each side of a point, which often occurs at points of discontinuity or where the function's graph has a sharp turn. It helps us understand the behavior of the function exclusively from the positive side of \(c\), without interference from the function's values when approached from the left.

When dealing with trigonometric functions like \(\cos x\) and \(\sec x\), approaching from the right has a specific significance. Since these functions are continuous and smooth around \(x = 0\), approaching from the right yields the same limit as if we approached from the left. So, in such a symmetrical context, the direction of the approach doesn’t change the limit value, and the statement 'as \(x \rightarrow 0^{+}\), \(\cos x\) and \(\sec x\) approach 1' holds true.