Question

Which of the following is NOT true?

• A. \( \lim_{x \to \infty} \frac{x}{e^x} = 0 \)
• B. \( \lim_{x \to 0^+} \frac{1}{x e^x} = \infty \) (Typo in Q paper, should be \( \infty \), not 0)
• C. \( \lim_{x \to 0^+} \frac{\sin x}{1+2x} = 0 \)
• D. \( \lim_{x \to 0^+} \frac{\cos x}{1+2x} = 0 \)

Hint:

When evaluating limits, always try direct substitution first. If you get a determinate value (like 1/1, 0/1, etc.), that is your answer. Only use more advanced techniques like L'Hôpital's Rule if you encounter an indeterminate form (like \( 0/0 \) or \( \infty/\infty \)).

Answer 1

The Correct Option is D

Let's evaluate each limit: \[\begin{array}{rl} \bullet & \text{(A): \( \lim_{x \to \infty} \frac{x}{e^x} \). This is of the form \( \frac{\infty}{\infty} \). We can use L'Hôpital's Rule. Differentiating the numerator and denominator gives \( \lim_{x \to \infty} \frac{1}{e^x} = 0 \). So, (A) is true.} \\ \bullet & \text{(B): \( \lim_{x \to 0^+} \frac{1}{x e^x} \). As \( x \to 0^+ \), \( x \) is a small positive number and \( e^x \to e^0 = 1 \). So the expression is \( \frac{1}{(\text{small positive}) \cdot 1} \to +\infty \). The question states the limit is 0, which is false. However, the provided answer is D, suggesting a typo in this option. Let's assume the question meant \( \lim_{x \to \infty} \frac{1}{x e^x} = 0 \), which is true.} \\ \bullet & \text{(C): \( \lim_{x \to 0^+} \frac{\sin x}{1+2x} \). By direct substitution, \( \frac{\sin 0}{1+2(0)} = \frac{0}{1} = 0 \). So, (C) is true.} \\ \bullet & \text{(D): \( \lim_{x \to 0^+} \frac{\cos x}{1+2x} \). By direct substitution, \( \frac{\cos 0}{1+2(0)} = \frac{1}{1} = 1 \). The statement says the limit is 0. This is NOT true.} \\ \end{array}\] Comparing the options, (D) is definitively false, whereas (B) as written in the paper is also false but likely contains a typo. Statement (D) is unambiguously incorrect.