Chapter 1: Problem 2
Find the discontinuities, if any. Evaluate (a) \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\) (b) \(\lim _{x \rightarrow 0} \frac{1-\cos x}{x}\).
Short Answer
Expert verified
Both functions have removable discontinuities at \( x = 0 \).
Step by step solution
01
Identifying the Function and Discontinuity for Part (a)
For part (a), we look at the function \( f(x) = \frac{\sin x}{x} \). This function is undefined at \( x = 0 \) due to division by zero, indicating a potential discontinuity at this point.
02
Evaluating the Limit for Part (a)
To evaluate \( \lim_{x \to 0} \frac{\sin x}{x} \), use the standard limit property that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). This result is derived from the definition of the derivative of \( \sin x \) at 0 or by using L'Hôpital's rule.
03
Conclusion for Part (a)
Since the limit as \( x \to 0 \) exists and equals 1, there is a removable discontinuity at \( x = 0 \) for \( \frac{\sin x}{x} \). If we redefine the function such that \( f(0) = 1 \), then the discontinuity is removed.
04
Identifying the Function and Discontinuity for Part (b)
For part (b), the function is \( g(x) = \frac{1 - \cos x}{x} \), which is also undefined at \( x = 0 \) because it involves division by zero. This again suggests a potential discontinuity at \( x = 0 \).
05
Evaluating the Limit for Part (b)
To find \( \lim_{x \to 0} \frac{1 - \cos x}{x} \), use L'Hôpital's rule because it's of the indeterminate form \( \frac{0}{0} \). Differentiating the numerator and denominator gives \( \lim_{x \to 0} \frac{\sin x}{1} \), which simplifies to \( 0 \).
06
Conclusion for Part (b)
The limit exists and is 0, indicating a removable discontinuity at \( x = 0 \) for \( \frac{1 - \cos x}{x} \). This means that if we define the function value at \( x = 0 \) to be 0, the discontinuity is removed.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits
Limits are a fundamental concept in calculus that help us understand the behavior of functions as they approach a certain point. Imagine limits as if you are zooming in on a function at a specific point to see what value it approaches, even if the function doesn't actually reach that value. For example:
- If you have the function \( f(x) = \frac{\sin x}{x} \), it is undefined at \( x = 0 \) because you cannot divide by zero.
- To find out what happens as \( x \) gets very close to zero, we calculate the limit: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). This means that as \( x \) approaches zero, the value of \( f(x) \) gets closer and closer to 1.
Discontinuities in functions
Discontinuities occur in functions where the graph has a jump, a break, or a hole. At these points, the function may be undefined or not behave as expected. Typically, discontinuities are classified into different types: removable, jump, and infinite discontinuities. Let’s focus on:
- Removable Discontinuities: These happen when a function is not defined at a point, but if the function could be defined at that point, it removes the break. It's like there's a small hole in the graph that we can fill by redefining the function at that specific point. A common example is \( f(x) = \frac{\sin x}{x} \) where at \( x = 0 \), if we set \( f(0) = 1 \), the function becomes continuous.
- Jump Discontinuities: Occur when there's a sudden change in the value of the function. The function jumps from one value to another.
- Infinite Discontinuities: These discontinuities occur where the function goes to infinity or negative infinity.
Removable discontinuities
Removable discontinuities are interesting because they occur when a function has a point of discontinuity that can be effectively "removed" by redefining the function at that specific point. Consider it like "filling in" a hole in the graph of the function.
- In mathematical terms, if a function \( f(x) \) has a limit \( L \) as \( x \) approaches \( c \), but \( f(c) \) is not defined or not equal to \( L \), it is a removable discontinuity.
- For example, in the case of \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), the function can be redefined to make it continuous if we define \( f(0) = 1 \).
- Another example is \( \frac{1 - \cos x}{x} \). At \( x = 0 \), this is undefined, but the limit is 0. By defining the function such that \( g(0) = 0 \), the discontinuity is removed.