Question:

Evaluate the limit: \[ \lim_{x \to \infty} \frac{A + e^{mx}}{x + Ae^{mx}} \]

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When exponential terms decay (as \( x \to -\infty \)), dominant terms in the numerator and denominator decide the limit.
Updated On: May 17, 2025
  • \( \frac{A}{x},\ x<0 \)
  • 1 when \( x>0 \)
  • 0 for all \( x \in \mathbb{R} \)
  • \( A,\ x = 0 \)
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The Correct Option is A

Solution and Explanation

As \( x \to \infty \), for \( x<0 \Rightarrow mx<0 \), so: \[ \begin{align} e^{mx} \to 0,\ \text{so } \lim_{x \to \infty} \frac{A + e^{mx}}{x + Ae^{mx}} \approx \frac{A}{x} \Rightarrow \text{expression tends to 0 as } x \to -\infty \] But since the question focuses on \( x<0 \), the dominant behavior is: \[ \lim_{x \to -\infty} \frac{A}{x} = 0^- \] Hence, the limit is effectively \( \frac{A}{x} \) in the region \( x<0 \)
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