Question

\[ \mathop {\lim }\limits_{x \to 0} \frac{\mathrm{e}^{x^2}-\cos 3 x}{\sin x \log (1+2 x)}= \]

• A. \(\frac{3}{2}\)
• B. \(\frac{-3}{2}\)
• C. \(\frac{11}{4}\)
• D. \(\frac{-11}{4}\)

Hint:

Expand the exponential and cosine near zero and keep only the leading terms in the numerator and denominator.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Both numerator and denominator tend to 0 as \(x\to0\), so we use the standard expansions for small \(x\).

Step 2: Numerator
\(e^{x^{2}}\approx 1+x^{2}\) and \(\cos 3x\approx 1-\dfrac{9x^{2}}{2}\).
\(e^{x^{2}}-\cos3x\approx x^{2}+\dfrac{9x^{2}}{2}=\dfrac{11x^{2}}{2}\)

Step 3: Denominator
\(\sin x\approx x\) and \(\log(1+2x)\approx 2x\), so the denominator is about \(2x^{2}\).

Step 4: Calculation
The limit is \(\dfrac{11x^{2}/2}{2x^{2}}=\dfrac{11}{4}\).

Final Answer:
The limit equals \(\frac{11}{4}\). \[ \boxed{\text{(C) }\dfrac{11}{4}} \]