Question

If \(A\neq 0\) and \(x>0\), then \[ \lim_{n\to\infty} \frac{\cos x-e^{-nx}} {1-Ae^{-nx}} = \]

• A. Does not exist
• B. \(1\)
• C. \(\dfrac{\cos x}{A}\)
• D. \(\dfrac{1}{A}\)

Hint:

When exponential terms like \(e^{-nx}\) occur with \(x>0\), first note that \(e^{-nx}\to0\). Then evaluate the limit directly or divide by the dominant exponential term if needed.

Answer 1

The Correct Option is D

Step 1: Evaluate the exponential term as n → ∞.
For x>0, nx → ∞, so e^(-nx) → 0.

Step 2: Rewrite by multiplying numerator and denominator by e^(nx).
L = lim_{n→∞} (cos x - e^(-nx))/(1 - Ae^(-nx)) = lim_{n→∞} (cos x·e^(nx) - 1)/(e^(nx) - A).

Step 3: Identify the dominant terms.
As n → ∞, e^(nx) dominates constants. The leading terms are cos x·e^(nx) in the numerator and e^(nx) in the denominator, so L = cos x.

Step 4: Compare with the given options.
The direct limit is cos x, though this may not appear among the printed choices, suggesting a possible typographical issue in the problem statement.

Step 5: Final conclusion.
The limit evaluates to cos x for the expression exactly as written.