Question

Determine the interval(s) on which the following functions are continuous; then evaluate the given limits. $$f(x)=\frac{e^{x}}{1-e^{x}} ; \lim _{x \rightarrow 0^{-}} f(x) ; \lim _{x \rightarrow 0^{+}} f(x)$$

Short answer

Question: Determine the intervals of continuity for the function $$f(x) = \frac{e^x}{1-e^x}$$ and evaluate the left-hand limit when $$x\rightarrow0^-$$ and the right-hand limit when $$x\rightarrow0^+$$. Answer: The function $$f(x)=\frac{e^{x}}{1-e^{x}}$$ is continuous on the intervals \((-\infty, 0) \cup (0, \infty)\). The left-hand limit when $$x\rightarrow0^-$$ and the right-hand limit when $$x\rightarrow0^+$$ do not exist.

Step-by-step solution

Identify the domain where the function is continuous

$$f(x)=\frac{e^{x}}{1-e^{x}}$$ is continuous wherever its denominator is not equal to zero, since division by zero is undefined. So, we need to find where $$1 - e^x \neq 0$$. This is equivalent to solving the equation $$1 - e^x = 0$$ for x. Solving the equation to find where the function is discontinuous: $$1-e^x=0$$ $$e^x=1$$ Taking the natural logarithm of both sides, we get: $$x=\ln(1)=0$$ Thus, the function is discontinuous at x=0. Therefore, the function is continuous on the intervals \((-\infty, 0) \cup (0, \infty)\).

Evaluate the left-hand limit

Now, let's evaluate the limit as $$x\rightarrow0^-$$. We are interested in the behavior of the function when x approaches 0 from the left side. As x approaches 0 from the left, the value of $$e^x$$ approaches 1. Thus, we have: $$\lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} \frac{e^{x}}{1-e^{x}}$$ Since e^x is continuous and approaches 1 in this case, we can evaluate the limit as: $$\lim _{x \rightarrow 0^{-}} \frac{e^{x}}{1-e^{x}} = \frac{1}{1-1}$$ However, as we noted earlier, the denominator is 0 at this point, and thus the limit does not exist.

Evaluate the right-hand limit

Now, let's evaluate the limit as $$x\rightarrow0^+$$. We want to know the behavior of the function when x approaches 0 from the right side. As x approaches 0 from the right, the value of $$e^x$$ also approaches 1. Thus, we have: $$\lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} \frac{e^{x}}{1-e^{x}}$$ Since e^x is continuous and approaches 1 in this case, we can evaluate the limit as: $$\lim _{x \rightarrow 0^{+}} \frac{e^{x}}{1-e^{x}} =\frac{1}{1-1}$$ Again, as we noted earlier, the denominator is 0 at this point, and thus the limit does not exist. In conclusion, the function $$f(x)=\frac{e^{x}}{1-e^{x}}$$ is continuous on the intervals \((-\infty, 0) \cup (0, \infty)\). The left-hand limit when $$x\rightarrow0^-$$ and the right-hand limit when $$x\rightarrow0^+$$ do not exist.

Key concepts

Limits of Functions

In mathematics, a limit describes how a function behaves as its argument approaches a specific point. Limits are crucial for determining the continuity and differentiability of functions.
The concept allows us to predict the behavior of functions near critical points, even when they might not be explicitly well-defined at those points.
To evaluate limits, especially near points of potential discontinuity, we often consider both left-hand (\( \lim_{x \rightarrow a^-} \)) and right-hand limits (\( \lim_{x \rightarrow a^+} \)).

• If both of these limits are equal at a point, the overall limit exists.
• If they are not equal, the limit is said to be undefined at that point.

In our exercise, the function \( f(x) = \frac{e^x}{1-e^x} \) has undefined limits as \( x \) approaches 0 from both directions.
This arises because the denominator, \( 1-e^x \), becomes 0, leading to undefined behavior.

Discontinuous Points

Discontinuous points in functions occur where a function is not continuous, typically caused by issues such as division by zero or jumps in graph behavior.
A function is continuous at a point if there's no interruption or hole in its graph.
It means that small changes in the input lead to small changes in the output.In the context of our function \( f(x) = \frac{e^x}{1-e^x} \), we identified that it is discontinuous at \( x = 0 \).

• To find discontinuities, we solve \( 1 - e^x = 0 \), leading us to \( x = \ln(1) = 0 \).
• This discontinuity arises because the denominator becomes zero, which is undefined.

Thus, the function is continuous over the intervals \( (-\infty, 0) \cup (0, \infty) \), avoiding the point \( x = 0 \).

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \) is approximately 2.71828.
It is a fundamental function in calculus, often used for solving equations involving exponential functions, differentiating products and quotients inside integrals, and more.
In logarithms, \( \ln(1) \) equals 0, since \( e^0 \) equals 1.
This property is vital when dealing with exponential equations.
Taking the natural logarithm of both sides of an equation like \( e^x = 1 \) allows us to isolate \( x \):

• This is precisely what we did in our problem to determine the point of discontinuity at \( x = 0 \).
• This knowledge helps solve various calculus-related problems, especially involving growth and decay processes.

Exponential Function

An exponential function is a mathematical expression in which a constant base is raised to a variable exponent, typically written as \( b^x \).
The standard exponential function is written with base \( e \), as \( e^x \). This function is inherently connected to processes involving constant percentage growth or decay over a time interval.
Exponential functions exhibit rapid growth or decay, depending on whether the exponent is positive or negative.
Some characteristics include:

• They are continuous and smooth for all real numbers.
• They always have an increasing slope if the base is above 1.

In our function \( f(x) = \frac{e^x}{1-e^x} \), \( e^x \) is always positive and approaches 1 as \( x \) approaches 0.
Understanding properties of exponential functions is key to solving calculus problems involving continuous and increasing or decreasing trends.