Question

lim_{x\to\infty} \frac{e^{3x} + 4}{e^{2x} - e^x}

Explanation:

Step1: Divide numerator and denominator by \( e^{3x} \)

To simplify the limit as \( x \to \infty \), we divide both the numerator and the denominator by the highest power of \( e^x \) in the denominator, which is \( e^{3x} \) (since \( e^{3x} \) grows faster than \( e^{2x} \) and \( e^x \) as \( x \to \infty \)).

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Step2: Simplify each term

Simplify the exponents using the rule \( \frac{e^a}{e^b} = e^{a - b} \):

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Step3: Evaluate the limit as \( x \to \infty \)

As \( x \to \infty \), \( e^{-x} = \frac{1}{e^x} \to 0 \), \( e^{-2x} = \frac{1}{e^{2x}} \to 0 \), and \( \frac{4}{e^{3x}} \to 0 \) because exponential functions with negative exponents approach 0 as \( x \to \infty \).

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Wait, that can't be right. Wait, I made a mistake in Step 1. The highest power in the numerator is \( e^{3x} \), so we should divide numerator and denominator by \( e^{3x} \), but let's re-examine.

Wait, the numerator is \( e^{3x} + 4 \), denominator is \( e^{2x} - e^x \). So when \( x \to \infty \), the dominant term in the numerator is \( e^{3x} \), and in the denominator, the dominant term is \( e^{2x} \)? No, wait, \( e^{3x} \) grows faster than \( e^{2x} \), so actually, the numerator grows faster than the denominator. Wait, maybe I should divide by \( e^{3x} \) in numerator and denominator.

Wait, let's redo Step 1:

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Now, as \( x \to \infty \), \( e^x \to \infty \), \( \frac{4}{e^{3x}} \to 0 \), and \( e^{-x} \to 0 \). So the numerator \( e^x(1 + 0) \to \infty \), and the denominator \( 1 - 0 = 1 \). So the limit is \( \infty \).

Wait, let's check again. The original limit:

Numerator: \( e^{3x} + 4 \), as \( x \to \infty \), \( e^{3x} \) dominates, so numerator ~ \( e^{3x} \).

Denominator: \( e^{2x} - e^x \), as \( x \to \infty \), \( e^{2x} \) dominates, so denominator ~ \( e^{2x} \).

So the limit is \( \lim_{x \to \infty} \frac{e^{3x}}{e^{2x}} = \lim_{x \to \infty} e^x = \infty \).

Ah, I see, my first mistake was in the division. Let's correct it.

Step1: Identify dominant terms

As \( x \to \infty \), \( e^{3x} \) (in numerator) and \( e^{2x} \) (in denominator) are the dominant terms (since they grow faster than the other terms in their respective numerator and denominator).

Step2: Approximate numerator and denominator

Approximate the numerator and denominator by their dominant terms:

$$ \lim_{x \to \infty} \frac{e^{3x} + 4}{e^{2x} - e^x} \approx \lim_{x \to \infty} \frac{e^{3x}}{e^{2x}} $$

Step3: Simplify the expression

Simplify \( \frac{e^{3x}}{e^{2x}} = e^{3x - 2x} = e^x \).

Step4: Evaluate the limit

As \( x \to \infty \), \( e^x \to \infty \).

So the limit is \( \infty \).

Answer:

\(\infty\) (or we can say the limit does not exist (diverges to infinity))