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The function given is:

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

We are tasked with evaluating the following limits:

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

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

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

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

At \( x = 0 \):
\[
f(0) = \frac{7e^0 + 6e^{0}}{e^{0} - e^0} = \frac{7 + 6}{1 - 1} = \frac{13}{0}
\]

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

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

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

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

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

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

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

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

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

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

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

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

### Final Results:

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

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

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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 \( x \to \infty \)?

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

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.

Learn how to evaluate the limits of the function f(x) = (7e^x + 6e^(4x)) / (e^(3x) - e^(4x)) as x approaches 0 from the left and right, and infinity. Understand the behavior of exponential functions, vertical asymptotes, and asymptotic limits in this college-level calculus problem.

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