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

Answer: The function is continuous on the intervals $$(-\infty,0)$$ and $$(0,\infty)$$. The limits are $$\lim _{x \rightarrow 0^{-}} f(x) = -\infty$$ and $$\lim _{x \rightarrow 0^{+}} f(x) = \infty$$.

Step-by-step solution

Identify Possible Points of Discontinuity

The function is not continuous when the denominator becomes zero. Let's find values of x for which $$1-e^x = 0$$. $$1-e^x = 0$$ $$e^x=1$$ The exponential function has only one solution to this equation, which is $$x=0$$.

Find Intervals of Continuity

Since the function has only one point of discontinuity at x=0, the intervals of continuity are $$(-\infty,0)$$ and $$(0,\infty)$$.

Find Left Limit at x=0

Now, let's find $$\lim _{x \rightarrow 0^{-}} f(x)$$, which means the limit as x approaches 0 from the left. $$\lim _{x \rightarrow 0^{-}} \frac{e^x}{1-e^x}$$ As x approaches 0 from the left, $$e^x$$ becomes very close to 1, but it never reaches 1. Hence, the denominator approaches 0, and the limit will be $$-\infty$$. So, $$\lim _{x \rightarrow 0^{-}} f(x) = -\infty$$.

Find Right Limit at x=0

Now, let's find $$\lim _{x \rightarrow 0^{+}} f(x)$$, which means the limit as x approaches 0 from the right. $$\lim _{x \rightarrow 0^{+}} \frac{e^x}{1-e^x}$$ As x approaches 0 from the right, $$e^x$$ becomes very close to 1 again but not equal to 1. The denominator approaches 0, and the limit will be $$\infty$$. So, $$\lim _{x \rightarrow 0^{+}} f(x) = \infty$$.

Final Answer

The given function $$f(x)=\frac{e^{x}}{1-e^{x}}$$ is continuous on the intervals $$(-\infty,0)$$ and $$(0,\infty)$$. The limits are as follows: $$\lim _{x \rightarrow 0^{-}} f(x) = -\infty$$ $$\lim _{x \rightarrow 0^{+}} f(x) = \infty$$

Key concepts

Limits of Functions

When studying functions, understanding the concept of limits is crucial. A limit helps us describe the behavior of a function as the input approaches a certain value. It captures how a function behaves near and at specific points, giving us valuable insight.

For instance, when we say \(\lim_{x \to 0} f(x) = L\), it means that as \(x\) gets closer and closer to 0, \(f(x)\) approaches the value \(L\). Finding limits can show us the function's behavior around points where it might not be defined, like our function \(f(x) = \frac{e^x}{1 - e^x}\) at \(x=0\).

Limiting behavior is crucial in calculus and analysis because it helps us investigate the function's continuity, differentiability, and possible points of discontinuity.

Points of Discontinuity

A point of discontinuity is where a function fails to be continuous. This can happen when a function jumps, breaks, or behaves erratically. In our exercise, we're interested in finding where the denominator \(1 - e^x\) of the function \(f(x)=\frac{e^{x}}{1-e^{x}}\) becomes zero. Solving \(1 - e^x = 0\) gives us \(e^x = 1\), leading to \(x = 0\).

If there’s any point where the function cannot be calculated in the usual sense, it's likely a discontinuity. At these points, the function may not exist, or its limit may not match the function's value if one exists.

Discontinuities are significant because they tell us where a function might not behave nicely, helping us partition the domain into intervals where the function can be analyzed as continuous, like \((-\infty, 0)\) and \((0, \infty)\) in this case.

Intervals of Continuity

Intervals of continuity are segments in the domain where a function behaves predictably and without any breaks. For the function \(f(x)=\frac{e^{x}}{1-e^{x}}\), it's continuous everywhere except at \(x=0\). We break the domain into continuous intervals: \((-\infty, 0)\) and \((0, \infty)\).

Within these intervals, you can calculate \(f(x)\) without running into division by zero or undefined expressions. For any point \(c\) in these intervals, \(f(x)\) should meet the condition: \(\lim_{x \to c} f(x) = f(c)\), meaning the function's limit at \(c\) equals the function's value at \(c\).

These continuous segments are useful for calculus operations, like integration and differentiation, where functions need to behave consistently within the interval.

One-Sided Limits

One-sided limits look at the behavior of functions as the variable approaches a point from one specific side, left or right. These are important when analyzing points of discontinuity. Take, for example, when \(x\) approaches 0 for our function \(f(x)=\frac{e^{x}}{1-e^{x}}\).

- **Left-hand limit**: \(\lim_{x \to 0^-} f(x)\) gives insight as \(x\) comes from the negative side. For this function, as \(x\) approaches 0 from the negative, \(f(x)\) tends toward \(-\infty\).- **Right-hand limit**: \(\lim_{x \to 0^+} f(x)\) focuses on \(x\) approaching from the positive side, suggesting \(f(x)\) soars to \(\infty\).

The fact that these two limits differ indicates a discontinuity at \(x=0\). Understanding one-sided limits is critical in confirming whether a function is truly continuous at a point, as both sides must harmonize for continuity.