Math Problem Statement

Determine the interval(s) on which the following function is continuous. Then analyze the given limits. \[ f(x) = \frac{3^x}{1 - e^{-4x}} \]

Solution

The given function is:

f(x)=3x1e4xf(x) = \frac{3^x}{1 - e^{-4x}}

1. Determining Continuity:

To determine where the function is continuous, we need to identify where it is defined. This function has a potential issue at points where the denominator becomes zero since division by zero is undefined.

The denominator 1e4x1 - e^{-4x} is zero when:

e4x=1e^{-4x} = 1 4x=0-4x = 0 x=0x = 0

Thus, x=0x = 0 is a point of discontinuity. We also need to check the behavior of 3x3^x and e4xe^{-4x}, but since exponential functions are defined for all real numbers, the only issue is at x=0x = 0 from the denominator.

Conclusion: f(x)f(x) is continuous on (,0)(0,)(-\infty, 0) \cup (0, \infty).

2. Analyzing the Limits:

Limit as xx approaches 0 from the left (x0x \to 0^-):

As xx approaches 0 from the left, e4xe^{-4x} becomes very large since 4x-4x becomes a large positive number. This makes 1e4x1 - e^{-4x} approach a large negative number, and since 3x3^x approaches 1 (as exponential functions are continuous), we have:

limx0f(x)=limx03x1e4x=1=0\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{3^x}{1 - e^{-4x}} = \frac{1}{-\infty} = 0

Limit as xx approaches 0 from the right (x0+x \to 0^+):

As xx approaches 0 from the right, e4xe^{-4x} becomes very small since 4x-4x becomes a large negative number. Therefore, 1e4x1 - e^{-4x} approaches 1, and 3x3^x also approaches 1:

limx0+f(x)=limx0+3x1e4x=11=1\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{3^x}{1 - e^{-4x}} = \frac{1}{1} = 1

Summary:

The function is continuous on (,0)(0,)(-∞, 0) \cup (0, ∞). The left-hand limit as xx approaches 0 is 0, and the right-hand limit as xx approaches 0 is 1.

Would you like more details on this or any other questions?

Some questions you might consider:

  1. How would the graph of this function look near x=0x = 0?
  2. What is the behavior of the function as xx approaches negative or positive infinity?
  3. How would the addition of a constant term in the denominator affect the limits?
  4. Can we differentiate this function to find its critical points?
  5. What is the significance of the base of the exponent (3 and -4) in this function's behavior?

Tip: When analyzing functions involving exponents and their limits, considering their behavior around points of discontinuity can often reveal interesting properties about the function’s overall behavior.

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Math Problem Analysis

Mathematical Concepts

Continuity
Limits
Exponential Functions

Formulas

\[ \lim_{x \to 0^-} f(x) = \frac{3^x}{1 - e^{-4x}} \]
\[ \lim_{x \to 0^+} f(x) = \frac{3^x}{1 - e^{-4x}} \]
Discontinuity occurs when the denominator is zero

Theorems

Continuity at a point
Limit theorems

Suitable Grade Level

Undergraduate Calculus