Problem 129 - Precalculus with Limits

Question

In Exercises 129 - 132, fill in the blanks. (Note:The notation \( x \rightarrow c^+ \) indicates that \( x \) approaches \( c \) from the right and \( x \rightarrow c^- \) indicates that \( x \) approaches \( c \) from the left.) As \( x \rightarrow \dfrac{\pi^-}{2} \), \( \sin x \rightarrow \square \) and \( \csc x \rightarrow \square \). .

Step-by-Step Solution

Answer

As \( x \rightarrow \dfrac{\pi^-}{2} \), \( \sin x \rightarrow 1 \) and \( \csc x \rightarrow 1 \).

Step 1: Approach of the limit

The exercise involves the calculation of a limit where \( x \rightarrow \dfrac{\pi^-}{2} \). This means that x is approaching \(\dfrac{\pi}{2}\) from left.

Step 2: Calculate the limit for sine function

When \( x \rightarrow \dfrac{\pi^-}{2} \), this means that x is getting very close to \(\dfrac{\pi}{2}\) from left. Substituting \(\dfrac{\pi}{2}\) into sin(x) gives sin(\(\dfrac{\pi}{2}\)), which equals to 1. So, as \( x \rightarrow \dfrac{\pi^-}{2} \), \( \sin x \rightarrow 1 \). This value fills the first blank in the exercise.

Step 3: Calculate the limit for cosecant function

The cosecant function, csc(x), is the reciprocal of the sine function. Therefore, when x is approaching \(\dfrac{\pi}{2}\) from the left and the value of sin(x) reaches its maximum 1, csc(x) will reach its minimum. Since csc(x) = 1/sin(x), csc(\(\dfrac{\pi}{2}\)) = 1/1, which equals to 1. So, as \( x \rightarrow \dfrac{\pi^-}{2} \), \( \csc x \rightarrow 1 \). This value fills the second blank.

Key Concepts

Approaching from the Left

When dealing with limits, the direction from which a variable approaches a point can significantly impact the outcome. Approaching from the left, denoted as \( x \to c^- \), means moving towards \( c \) from values that are less than \( c \).
This is opposed to approaching from the right, which involves moving towards \( c \) from values that are greater than \( c \).
Understanding this notation is crucial in limit calculations:

• Limits from the left are often associated with approaching numbers smaller than the target value.
• This approach helps in identifying the behavior of a function just before it reaches a particular point.

In our exercise, \( x \rightarrow \frac{\pi^-}{2} \) indicates getting close to \( \frac{\pi}{2} \) while remaining on the lesser side of the interval.
This distinction is essential to accurately predict the limit value of the functions involved.

Sine Function

The sine function, often written as \( \sin x \), is a fundamental trigonometric function that relates an angle in a circle to the ratio of the length of the opposite side to the hypotenuse in a right triangle.
As \( x \) increases and approaches \( \frac{\pi}{2} \) from the left (or from smaller values), the sine function nears its peak value.
Here's a breakdown of how this works:

• \( \sin(\frac{\pi}{2}) = 1 \); it is where the sine function reaches its maximum value.
• As \( x \to \frac{\pi^-}{2} \), \( \sin x \to 1 \)

This growth towards 1 indicates the highest point on the sine graph as \( x \) progresses from the left towards \( \frac{\pi}{2} \).This behavior is consistent with one of the core characteristics of the sine function, as it oscillates between -1 and 1.

Cosecant Function

The cosecant function is the reciprocal of the sine function, defined as \( \csc x = \frac{1}{\sin x} \).
Since it reverses the sine function's output, its behavior significantly differs, especially near critical points.
Understanding \( \csc x \) near \( \frac{\pi}{2} \):

• As \( x \to \frac{\pi^-}{2} \), \( \sin x \to 1 \).
• Thus, \( \csc x = \frac{1}{\sin x} \to \frac{1}{1} = 1 \).

This outcome showcases how \( \csc x \) reaches its maximum logical boundary within its range as \( x \) closes in from the left to \( \frac{\pi}{2} \).
While \( \csc x \) can exhibit undefined behavior when \( \sin x \) is zero, in this particular case, it resolves neatly to 1, mirroring the integrity of its sine counterpart at this point.