Question

Determine the interval(s) on which the following functions are continuous; then analyze 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 find the left-hand limit and right-hand limit as $$x$$ approaches 0. Answer: The function is continuous on the intervals $$(-\infty, 0)$$, and $$(0, \infty)$$. As $$x$$ approaches 0 from the left, the left-hand limit is $$+\infty$$, and as $$x$$ approaches 0 from the right, the right-hand limit is $$-\infty$$.

Step-by-step solution

Determine the intervals of continuity

First, we need to find the intervals where the function $$f(x)$$ is continuous. A rational function is continuous wherever its denominator is non-zero. Thus, we need to examine the denominator and find where it is not equal to zero. Denominator: $$1 - e^{x}$$ Let's set the denominator equal to zero and solve for $$x$$: $$1 - e^{x} = 0 \Rightarrow e^{x} = 1$$ Taking the natural logarithm of both sides, we get: $$x = \ln 1 = 0$$ The denominator is zero at $$x = 0$$. Therefore, the function is continuous on the intervals $$(-\infty, 0)$$ and $$(0, \infty)$$.

Find the left-hand limit

Now, we need to find the left-hand limit of the function as $$x$$ approaches 0: $$\lim _{x \rightarrow 0^{-}} f(x)$$ Here, we need to determine the behavior of the function as $$x$$ approaches 0 from the left (i.e., the negative side). As $$x$$ approaches 0 from the left, $$e^{x}$$ approaches 1, but remains less than 1. Therefore, the denominator also approaches 0, but remains positive: $$1 - e^{x} \rightarrow 0^{+}$$ Consequently, the function approaches the value: $$f(x) = \frac{e^{x}}{1 - e^{x}}$$ Since the numerator approaches 1 and the denominator approaches 0 from the positive side, the left-hand limit is: $$\lim _{x \rightarrow 0^{-}} f(x) = \frac{1}{0^{+}} = +\infty$$

Find the right-hand limit

Similarly, we need to find the right-hand limit of the function as $$x$$ approaches 0: $$\lim _{x \rightarrow 0^{+}} f(x)$$ As $$x$$ approaches 0 from the right, $$e^{x}$$ approaches 1, but remains greater than 1. Therefore, the denominator approaches 0, but remains negative: $$1 - e^{x} \rightarrow 0^{-}$$ The function will then approach the value: $$f(x) = \frac{e^{x}}{1 - e^{x}}$$ Since the numerator approaches 1 and the denominator approaches 0 from the negative side, the right-hand limit is: $$\lim _{x \rightarrow 0^{+}} f(x) = \frac{1}{0^{-}} = -\infty$$ In conclusion, the function $$f(x) = \frac{e^{x}}{1 - e^{x}}$$ is continuous on the intervals $$(-\infty, 0)$$, and $$(0, \infty)$$. When approaching 0 from the left, the function tends to $$+\infty$$, and when approaching 0 from the right, it tends to $$-\infty$$.

Key concepts

Continuity in Calculus

Understanding continuity in calculus is vital for analyzing how functions behave. A function is said to be continuous at a point if the function exists at that point, the limit of the function as it approaches the point from both sides exists, and both the function value and the limit are equal. In simpler terms, you can think of it as being able to draw the graph of the function without lifting your pencil.

For rational functions, which are ratios of two polynomials, continuity primarily depends on the denominator not being equal to zero since division by zero is undefined. In our exercise with the function \( f(x) = \frac{e^{x}}{1 - e^{x}} \), we determined continuity by first identifying where the denominator does not equal zero, which led to intervals \( (-\infty, 0) \) and \( (0, \infty) \).

Exploring Intervals of Continuity

By solving \( 1 - e^{x} = 0 \), we found that \( x = 0 \) is where the function is not continuous, hence \( f(x) \) is continuous everywhere except at \( x = 0 \).

Rational Functions

Rational functions are expressions formed by the ratio of two polynomials, such as \( \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials. The key to understanding rational functions lies in their domains, which are all real numbers except for where the denominator \( q(x) \) equals zero.

In the provided exercise, our rational function has an exponential function in the denominator. This slightly deviates from typical polynomials but the principle remains the same. The denominator, \( 1 - e^{x} \) in our exercise, sets the stage for the function's domain and points of discontinuity.

Denominator's Impact on Domain

Since the function cannot exist where the denominator is zero, identifying these points is crucial for determining the function's domain and its continuity, just as we did by solving \( 1 - e^{x} = 0 \).

Left-Hand Limit

A left-hand limit examines the behavior of a function as the input approaches a certain value from the left side, or from values smaller than the point of interest. Symbolically, it's represented as \( \lim_{{x \to c^{-}}} f(x) \), meaning 'the limit of \( f(x) \) as \( x \) approaches \( c \) from the left'.

In our exercise, finding the left-hand limit at \( x = 0 \) for the function \( f(x) \) involved understanding the behavior of \( e^{x} \) as \( x \) gets closer and closer to zero from negative values. The value of \( e^{x} \) nears 1, making the denominator very small and positive, resulting in a large positive value for \( f(x) \) itself, hence \( +\infty \) as the left-hand limit.

Right-Hand Limit

Conversely, a right-hand limit investigates the function's behavior as the input value comes from the right, or from values larger than the point of interest. This is denoted as \( \lim_{{x \to c^{+}}} f(x) \) and means 'the limit of \( f(x) \) as \( x \) approaches \( c \) from the right'.

In our function \( f(x) \) as \( x \) approaches 0 from the right, the resulting right-hand limit involves the same denominator nearing zero but from negative values this time, due to \( e^{x} \) being just slightly over 1. This tiny negative denominator thus leads to the function \( f(x) \) spiraling down to \( -\infty \) as the right-hand limit.

Asymptotic Behavior

Asymptotic behavior refers to how a function behaves as it moves towards infinity or towards a certain point where the function may not be defined - typically a point of discontinuity. In graphs, asymptotes are lines that the curve approaches but never actually reaches.

In the context of our exercise, as \( x \) approaches 0, the function \( f(x) \) does not settle on a finite number but instead tends towards \( +\infty \) from the left and \( -\infty \) from the right. This is an indication of vertical asymptotes at \( x = 0 \) for the function \( f(x) \).

Understanding Vertical Asymptotes

Vertical asymptotes occur at points where the function shoots up to \( +\infty \) or down to \( -\infty \) as the input gets infinitely close to a certain value, in this case, \( x = 0 \) for our function.