Question

Classify the discontinuities in the following functions at the given points. See Exercises \(91-92.\) $$h(x)=\frac{x^{3}-4 x^{2}+4 x}{x(x-1)} ; x=0 \text { and } x=1$$

Short answer

Question: Classify the discontinuities of the function \(h(x)=\frac{x^{3}-4 x^{2}+4 x}{x(x-1)}\) at the points \(x=0\) and \(x=1\). Answer: The function \(h(x)\) has removable discontinuities at \(x=0\) and \(x=1\).

Step-by-step solution

Identify Discontinuities at \(x=0\)

To identify the type of discontinuity at \(x=0\), we will find the limit of the function as \(x\) approaches \(0\). Let's factor the numerator first: $$h(x)=\frac{x^3-4x^2+4x}{x(x-1)} = \frac{x(x^2-4x+4)}{x(x-1)}=\frac{x(x-2)^2}{x(x-1)}$$ Now, let's find the limit: $$\lim_{x \to 0} \frac{x(x-2)^2}{x(x-1)}$$ As \(x\) approaches \(0\), the limit approaches \(\frac{0}{0}\). This is an indeterminate form, and therefore we need to simplify the expression further before evaluating the limit. After simplifying, we get: $$\lim_{x \to 0} \frac{(x-2)^2}{x-1}$$ Now, we can evaluate the limit as \(x\) approaches \(0\): $$\lim_{x \to 0} \frac{(-2)^2}{-1} = 4$$ Since the limit of the function exists as \(x\) approaches \(0\), there is a removable discontinuity at \(x=0\).

Identify Discontinuities at \(x=1\)

To identify the type of discontinuity at \(x=1\), we will find the limit of the function as \(x\) approaches \(1\): $$\lim_{x \to 1} \frac{x(x-2)^2}{x(x-1)}$$ As \(x\) approaches \(1\), the limit approaches \(\frac{0}{0}\). This is an indeterminate form, and therefore we will simplify the expression again: $$\lim_{x \to 1} \frac{(x-2)^2}{x}$$ Now, we can evaluate the limit as \(x\) approaches \(1\): $$\lim_{x \to 1} \frac{(-1)^2}{1} = 1$$ Since the limit of the function exists as \(x\) approaches \(1\), there is a removable discontinuity at \(x=1\).

Conclusion

The function \(h(x)=\frac{x^{3}-4 x^{2}+4 x}{x(x-1)}\) has removable discontinuities at \(x=0\) and \(x=1\).

Key concepts

Removable Discontinuity

A removable discontinuity occurs in a function when there is a hole at a certain point on its graph. This happens when a function is not defined at that point, but you could redefine it so that it is continuous. Essentially, you could "fill in the hole" by appropriately defining the function at that point.

In the exercise given, the function \(h(x)\) has removable discontinuities at \(x=0\) and \(x=1\). This is because both points initially resulted in an indeterminate form of \(\frac{0}{0}\). After simplifying the function through factoring and canceling common terms, the function becomes continuous at these points if the function is properly defined by the limit result.

For instance:

• At \(x=0\), the simplified function and its limit result in a value of \(4\). By defining the function to equal \(4\) when \(x=0\), the discontinuity is removed.
• Similarly, at \(x=1\), the limit also exists, with the function equaling \(1\). Again, by redefining the function at \(x=1\) to equal \(1\), the discontinuity can be removed.

Limits

Limits are a fundamental concept in calculus that describe the behavior of a function as it approaches a particular point. Understanding limits allows us to deal with values at which a function is not explicitly defined, such as in cases of removable discontinuities.

In the context of the given exercise, calculating the limits of \(h(x)\) as \(x\) approaches \(0\) and \(1\) helps determine whether a discontinuity at these points is removable. When initially calculating the limit at these points, both resulted in \(\frac{0}{0}\), indicating an indeterminate form. However, by simplifying the expression, limits were successfully evaluated:

• When \(x\) approaches \(0\), after simplification, the limit was found to be \(4\).
• When \(x\) approaches \(1\), the limit was \(1\).

Recognizing and simplifying these limits allowed us to identify the removable nature of these discontinuities.

Indeterminate Forms

Indeterminate forms are expressions that do not lead to a clear, immediate result. In calculus, the most common indeterminate forms include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). These forms occur when evaluating a mathematical limit or a situation where both the numerator and denominator simultaneously approach zero.

In the initial steps of evaluating the function \(h(x)\) for \(x=0\) and \(x=1\), we encountered an indeterminate form of \(\frac{0}{0}\). This forms a challenge as it does not provide a direct answer for the limit.

• Addressing indeterminate forms requires simplifying the function, often by factor cancelation, which eliminates the term causing the zero in the denominator.
• Only after simplification were we able to compute the limits, confirming them as \(4\) and \(1\) for \(x=0\) and \(x=1\) respectively.

This process allowed us to recognize that the discontinuities at those points were removable.