Math Problem Statement

Evaluate limits for the function f(x) = (7e^x + 6e^(4x)) / (e^(3x) - e^(4x)) at x approaching 0 from the left, right, negative infinity, and positive infinity.

Solution

The function given is:

f(x)=7ex+6e4xe3xe4xf(x) = \frac{7e^x + 6e^{4x}}{e^{3x} - e^{4x}}

We are tasked with evaluating the following limits:

  1. limx0f(x)\lim_{x \to 0^-} f(x)
  2. limx0+f(x)\lim_{x \to 0^+} f(x)
  3. limxf(x)\lim_{x \to -\infty} f(x)
  4. limxf(x)\lim_{x \to \infty} f(x)

Let's calculate each of these limits step by step.

1. limx0f(x)\lim_{x \to 0^-} f(x)

When x0x \to 0, we evaluate the function by substituting values close to 0 from the left-hand side:

At x=0x = 0: f(0)=7e0+6e0e0e0=7+611=130f(0) = \frac{7e^0 + 6e^{0}}{e^{0} - e^0} = \frac{7 + 6}{1 - 1} = \frac{13}{0}

Since the denominator tends to 0, this results in a vertical asymptote. When approaching 0 from the negative side, the denominator e3xe4xe^{3x} - e^{4x} is negative (as e4xe^{4x} dominates). Thus:

limx0f(x)=+\lim_{x \to 0^-} f(x) = +\infty

2. limx0+f(x)\lim_{x \to 0^+} f(x)

Now approaching 0 from the right-hand side, we get the same numerator, but the denominator e3xe4xe^{3x} - e^{4x} is now negative since e4x>e3xe^{4x} > e^{3x}. Therefore:

limx0+f(x)=\lim_{x \to 0^+} f(x) = -\infty

3. limxf(x)\lim_{x \to -\infty} f(x)

As xx \to -\infty, both exe^x and e4xe^{4x} tend to 0 because exponentials with negative powers decay to zero. Thus, the numerator 7ex+6e4x7e^x + 6e^{4x} and denominator e3xe4xe^{3x} - e^{4x} both approach 0. However, e3xe^{3x} dominates in the denominator as it decays slower than e4xe^{4x}. Therefore:

limxf(x)=0\lim_{x \to -\infty} f(x) = 0

4. limxf(x)\lim_{x \to \infty} f(x)

As xx \to \infty, the exponential functions exe^x, e4xe^{4x}, and e3xe^{3x} grow rapidly. Here, e4xe^{4x} dominates both the numerator and denominator because it has the highest exponent. Thus:

f(x)6e4xe4x=6f(x) \approx \frac{6e^{4x}}{-e^{4x}} = -6

Therefore: limxf(x)=6\lim_{x \to \infty} f(x) = -6

Final Results:

  1. limx0f(x)=+\lim_{x \to 0^-} f(x) = +\infty
  2. limx0+f(x)=\lim_{x \to 0^+} f(x) = -\infty
  3. limxf(x)=0\lim_{x \to -\infty} f(x) = 0
  4. limxf(x)=6\lim_{x \to \infty} f(x) = -6

Let me know if you want any additional explanations or if you'd like to explore more details.


Here are 5 related questions to expand on the topic:

  1. How do you calculate horizontal asymptotes for functions with exponential terms?
  2. What role do exponential growth rates play in evaluating limits of functions?
  3. How do vertical asymptotes impact the behavior of a function near certain points?
  4. Can a function have different behaviors when approaching a point from the left and right?
  5. How does the dominance of exponential terms affect limits as xx \to \infty?

Tip: When evaluating limits at infinity, always check which terms dominate in both the numerator and denominator, especially in exponential functions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Exponential Functions
Asymptotic Behavior

Formulas

Limit formula: lim_{x->a} f(x)
Exponential growth: e^x
Asymptotic behavior of functions

Theorems

Limit Theorem
Behavior of exponential functions at infinity and zero
Vertical Asymptotes

Suitable Grade Level

College-level Calculus