Problem 41 In Problems $41-48,$ determine... [FREE SOLUTION]

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Chapter 1: Problem 41

In Problems $41-48,$ determine whether the function is continuous at the given point c. If the function is not continuous, determine whether the discontinuity is removable or nonremovable. $$ f(x)=\sin x ; c=0 $$

Short Answer

Expert verified

The function $ f(x) = \sin x $ is continuous at $ c = 0 $ with no discontinuities.

Step by step solution

Identify the function and point

The problem asks us to determine the continuity of the function $ f(x) = \sin x $ at the point $ c = 0 $.

Understand Continuity

A function is continuous at a point $ c $ if the following conditions are met: 1) $ f(c) $ is defined, 2) The limit $ \lim_{x \to c} f(x) $ exists, and 3) $ \lim_{x \to c} f(x) = f(c) $.

Check if $ f(x) $ is defined at $ c $

The function $ f(x) = \sin x $ is defined for all real numbers, including $ x = 0 $. Therefore, $ f(0) = \sin 0 = 0 $.

Evaluate the limit of $ f(x) $ as $ x \to 0 $

Find $ \lim_{x \to 0} \sin x $. As $ x $ approaches 0, $ \sin x $ approaches $ \sin 0 = 0 $. Thus, $ \lim_{x \to 0} \sin x = 0 $.

Compare limit with function value at $ c $

We found that $ f(0) = 0 $ and $ \lim_{x \to 0} \sin x = 0 $. Since the limit equals the function value, $ \sin x $ is continuous at $ x = 0 $.

Determine if there is any discontinuity

Since the function is continuous at $ x = 0 $, there are no discontinuities (removable or non-removable) at this point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sine Function

The sine function, denoted as $ \sin x $, is a fundamental trigonometric function that measures the y-coordinate of a point on the unit circle as the angle $ x $ varies. It's periodic with a period of $ 2\pi $, meaning it repeats its values every $ 2\pi $ radians. The range of the sine function is from -1 to 1, making it bounded. Additionally, it is defined for every real number $ x $, ensuring its widespread applicability in many mathematics problems. Hence, at any angle including $ x = 0 $, $ \sin x $ is well-defined, and as we will see, it plays a key role in determining continuity at certain points.

Limits

Limits are a core concept in calculus and analyze the behavior of functions as they approach a certain point. For a function $ f(x) $, the limit $ \lim_{x \to c} f(x) $ tells us what value $ f(x) $ approaches as $ x $ gets infinitely close to $ c $. In the case of the sine function at $ x = 0 $, we evaluate how $ \sin x $ behaves as it nears 0. Calculating $ \lim_{x \to 0} \sin x $, we find it equals $ \sin 0 = 0 $. Thus, the limit aligns perfectly with the actual function value at 0, which is a crucial factor in assessing the function's continuity.

Continuous Function

A continuous function is one that does not have any jumps, breaks, or holes at a given point. Formally, a function $ f(x) $ is continuous at a point $ c $ if three conditions are met:

$ f(c) $ is defined.
The limit $ \lim_{x \to c} f(x) $ exists.
The limit equals the function value at $ c $, i.e., $ \lim_{x \to c} f(x) = f(c) $.

For the sine function at $ x = 0 $, we confirmed that $ \sin 0 $ is defined, the limit exists, and both the limit and $ \sin(0) $ equal 0. This seamless alignment shows that the sine function is continuous at $ x = 0 $.

Removable Discontinuity

A removable discontinuity occurs in a function when there is a hole at a certain point that can often be "fixed" by re-defining the function value at that point. However, in the case of $ \sin x $ at $ x = 0 $, there is no hole or gap. The function is smooth and continuous, meaning there are no removable discontinuities. If there were a removable discontinuity, we would see a break or gap in the graph that could be corrected, but for $ \sin x $, this is not the case.

Nonremovable Discontinuity

Nonremovable discontinuities are more stubborn and cannot be eliminated by simple redefinition. They occur when the limits from the left-hand and right-hand sides of a function do not match or when a function goes to infinity at a point. In our sine function at $ x = 0 $, no such discrepancies exist. The left-hand and right-hand limits as $ x \to 0 $ match perfectly, and $ \sin x $ does not go to infinity. Hence, the sine function exhibits no nonremovable discontinuities at $ x = 0 $, maintaining its continuity without any breaks or infinite behaviors.

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91影视

In Problems \(41-48,\) determine whether the function is continuous at the given point c. If the function is not continuous, determine whether the discontinuity is removable or nonremovable. $$ f(x)=\sin x ; c=0 $$

Short Answer

Step by step solution

Identify the function and point

Understand Continuity

Check if \( f(x) \) is defined at \( c \)

Evaluate the limit of \( f(x) \) as \( x \to 0 \)

Compare limit with function value at \( c \)

Determine if there is any discontinuity

Key Concepts

Sine Function

Limits

Continuous Function

Removable Discontinuity

Nonremovable Discontinuity

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