Question

Refer to Exercises \(91-92.\) a. Does the function \(f(x)=x \sin (1 / x)\) have a removable discontinuity at \(x=0 ?\) b. Does the function \(g(x)=\sin (1 / x)\) have a removable discontinuity at \(x=0 ?\)

Short answer

Answer: \(f(x)=x \sin (1 / x)\) has a removable discontinuity, while \(g(x)=\sin (1 / x)\) does not have a removable discontinuity at \(x=0\).

Step-by-step solution

Analyze Function \(f(x) = x \sin(1/x)\)

We'll find the limit as \(x\) approaches \(0\) for this function: \(\displaystyle\lim_{x\to 0} x \sin(\frac{1}{x})\).

Apply Squeeze Theorem

To find the limit of \(f(x)\), we can use the Squeeze Theorem. The Squeeze Theorem states that if \(g(x) \leq f(x) \leq h(x)\), and \(\displaystyle\lim_{x\to a} g(x) = \displaystyle\lim_{x\to a} h(x) = L\), then \(\displaystyle\lim_{x\to a} f(x) = L\).

Find Bounds for \(f(x)\)

We know that \(-1 \leq \sin(\frac{1}{x}) \leq 1\). So, we can multiply all three sides of the inequality by \(x\) to get: \(-x \leq x \sin(\frac{1}{x}) \leq x\)

Calculate Limit for Bounds

Now, we'll calculate the limit as \(x\) approaches \(0\) for the two bounds: \(-x\) and \(x\). \(\displaystyle\lim_{x\to 0} -x = 0\) and \(\displaystyle\lim_{x\to 0} x = 0\) As both limits equal to \(0\), using the Squeeze Theorem, it implies that \(\displaystyle\lim_{x\to 0} x \sin(\frac{1}{x}) = 0\)

Determine Removable Discontinuity for \(f(x)\)

Because the limit exists when \(x\) approaches \(0\) in \(f(x)\), we can say that it has a removable discontinuity at \(x=0\).

Analyze Function \(g(x) = \sin(1/x)\)

Now, we'll repeat the process for \(g(x)\). We need to find the limit as \(x\) approaches \(0\) for this function: \(\displaystyle\lim_{x\to 0} \sin(\frac{1}{x})\).

Apply Squeeze Theorem once again

We can apply the Squeeze Theorem just like we did for \(f(x)\). As before, we know that \(-1 \leq \sin(\frac{1}{x}) \leq 1\). However, we can't find any bounds in terms of \(x\) that would help us find the limit.

Determine Removable Discontinuity for \(g(x)\)

Since we cannot find any appropriate bounds in terms of \(x\) that would help us find the limit of \(g(x)\) as \(x\) approaches \(0\), the limit does not exist. Thus, \(g(x)\) does not have a removable discontinuity at \(x=0\). To summarize, the function \(f(x)=x \sin (1 / x)\) has a removable discontinuity at \(x=0\), while the function \(g(x)=\sin (1 / x)\) does not have a removable discontinuity at \(x=0\).

Key concepts

Squeeze Theorem

The Squeeze Theorem is a clever technique used in calculus to find the limit of a function that is "squeezed" between two other functions. To apply the theorem, you need three functions:

• If you have a function \( f(x) \) that you’re trying to evaluate as \( x \) approaches a point \( a \), and you know that \( g(x) \leq f(x) \leq h(x) \) near \( a \), it sets up the squeeze.
• If both \( g(x) \) and \( h(x) \) have the same limit as \( x \rightarrow a \), say \( L \), then \( f(x) \) must also have that same limit, \( L \). This power lies in using known limits to "squeeze" the unknown function into clarity.

In our specific exercise, we used the Squeeze Theorem on the function \( f(x) = x \sin(1/x) \). The function \( \sin(1/x) \) oscillates between -1 and 1, allowing us to bind \( f(x) \) between \( -x \) and \( x \). Since both boundaries go to \( 0 \) as \( x \) approaches \( 0 \), the Squeeze Theorem tells us \( f(x) \) does too.

limits

Limits are a foundational concept in calculus, describing what a function approaches as the input approaches a specific point.

• The notation \( \lim_{x \to a} f(x) = L \) essentially means that as \( x \) gets closer and closer to \( a \), \( f(x) \) gets closer and closer to \( L \).
• Limits help us deal with functions at points where they might not be well-defined, like division by zero.

In the exercise example, we're looking at \( \lim_{x \to 0} x \sin(1/x) \) and \( \lim_{x \to 0} \sin(1/x) \). While \( f(x) \) smoothly approaches 0 thanks to the Squeeze Theorem, \( g(x) = \sin(1/x) \) does not settle at any particular value when \( x = 0 \) is reached. This leads us to different conclusions about removable discontinuity.

functions

Understanding functions is essential as it forms the basis of relationship mapping in mathematics. A function is essentially a "rule" that assigns each input exactly one output.

• In our case, both \( f(x) = x \sin(1/x) \) and \( g(x) = \sin(1/x) \) are functions with \( x \) as the input variable.
• The distinct behavior of these functions at \( x = 0 \) highlights diversity in function types.

When dissecting these functions, \( f(x) \) works well with limits because the multiplication by \( x \) dilutes the oscillation effect of \( \sin(1/x) \), effectively "calming" it down as it nears zero. However, \( g(x) \) does not have an \( x \) damping factor, so it remains turbulent and lacks a finite limit. Observing these core differences is crucial in analyzing removable discontinuities.

discontinuity analysis

Discontinuity analysis is a way to understand how and why functions don't behave in expected "smooth" manners. In calculus, a point of discontinuity is where a function is not continuous.

• A removable discontinuity occurs when a function has a limit at a point where it's not actually defined, but can be "patched" or "remedied".
• For \( f(x) = x \sin(1/x) \), the function reaches \( x = 0 \) without issue, implying a removable discontinuity where it can be fixed by just defining \( f(0) = 0 \).

However, \( g(x) = \sin(1/x) \) cannot be patched since it oscillates wildly as \( x \) approaches \( 0 \), meaning there's no single value it can be fixed to without altering the essential nature of the original function. Recognizing these patterns helps in deciding how or if one might "fix" continuity issues.