Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76.
Try it—you just keep on cycling. Find the limits some other way.
73. lim (x → ∞) (2^x - 3^x) / (3^x + 4^x)

검증된 단계별 안내

Identify the limit expression: $\lim_{x \to \infty} \frac{2^x - 3^x}{3^x + 4^x}$.

Notice that as $x$ approaches infinity, the terms with the highest exponential base will dominate both numerator and denominator. Here, compare the bases: 2, 3, and 4.

Divide both numerator and denominator by the largest exponential term in the denominator, which is \$4^x$, to simplify the expression:

\[\frac{\frac{2^x}{4^x} - \frac{3^x}{4^x}}{\frac{3^x}{4^x} + \frac{4^x}{4^x}} = \frac{\left(\frac{2}{4}\right)^x - \left(\frac{3}{4}\right)^x}{\left(\frac{3}{4}\right)^x + 1}.\]

Analyze the behavior of each term as $x \to \infty$: since $\frac{2}{4} = 0.5$ and $\frac{3}{4} = 0.75$, both bases are less than 1, so their powers tend to zero.

Conclude the limit by substituting the limits of these terms into the simplified expression and evaluating the resulting constant expression.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Limits at Infinity

Limits at infinity describe the behavior of a function as the input grows without bound. Understanding how exponential functions behave as x approaches infinity is crucial, especially since different bases grow at different rates, affecting the limit's value.

Dominant Term in Expressions

When evaluating limits involving sums or differences of exponential functions, the term with the largest base dominates as x approaches infinity. Identifying and factoring out this dominant term simplifies the expression and helps find the limit.

L’Hôpital’s Rule and Its Limitations

L’Hôpital’s Rule is used to evaluate indeterminate forms like 0/0 or ∞/∞ by differentiating numerator and denominator. However, in some cases, such as this problem, repeated application leads to cycling without resolution, indicating alternative methods are needed.

Theory and ApplicationsL’Hôpital’s Rule does not help with the limits in Exercises 69–76. Try it—you just keep on cycling. Find the limits some other way.73. lim (x → ∞) (2^x - 3^x) / (3^x + 4^x)