Question

Which of the following is FALSE?

• A. \( \lim_{x \to \infty} \dfrac{x}{e^x} = 0 \)
• B. \( \lim_{x \to 0^+} \dfrac{1}{x e^{1/x}} = 0 \)
• C. \( \lim_{x \to 0^+} \dfrac{\sin x}{1 + 2x} = 0 \)
• D. \( \lim_{x \to 0^+} \dfrac{\cos x}{1 + 2x} = 0 \)

Hint:

Always check small-angle approximations like \(\sin x \approx x\) and \(\cos x \approx 1\) for limit problems near 0.

Answer 1

The Correct Option is D

Step 1: Evaluate each limit.
(A) \( \lim_{x \to \infty} \dfrac{x}{e^x} = 0 \) (exponential dominates polynomial).
(B) \( \lim_{x \to 0^+} \dfrac{1}{x e^{1/x}} = 0 \) since \( e^{1/x} \) grows faster than \( \dfrac{1}{x} \).
(C) \( \lim_{x \to 0^+} \dfrac{\sin x}{1 + 2x} = 0 \) since \( \sin x \approx x \).
(D) \( \lim_{x \to 0^+} \dfrac{\cos x}{1 + 2x} = \dfrac{1}{1} = 1 \neq 0 \).

Step 2: Conclusion.
Option (D) is false because the limit is 1, not 0.