Question

Evaluate \(\lim_{x \to 0} \frac{x^2 \log(\cos x)}{\log(1+x^2)}\)

• A. 0
• B. 1
• C. -1
• D. \(\infty\)

Hint:

Use Taylor series expansions of \(\cos x\) and \(\log(1+x^2)\) to simplify limit expressions.

Answer 1

The Correct Option is A

Step 1: Use the standard limit results directly.
Recall two standard small-angle limits: \(\lim_{x \to 0} \frac{\log(\cos x)}{x^2} = -\frac{1}{2}\) and \(\lim_{x \to 0} \frac{\log(1+x^2)}{x^2} = 1\).

Step 2: Rewrite the given limit using these.
\[ \lim_{x \to 0} \frac{x^2 \log(\cos x)}{\log(1+x^2)} = \lim_{x \to 0} x^2 \cdot \frac{\log(\cos x)/x^2}{\log(1+x^2)/x^2} \]

Step 3: Substitute the standard values.
As \(x \to 0\), the fraction \(\frac{\log(\cos x)/x^2}{\log(1+x^2)/x^2}\) tends to \(\frac{-1/2}{1}=-\frac{1}{2}\), so the whole expression behaves like \(-\frac{x^2}{2}\), which goes to 0 as \(x \to 0\).

Step 4: Final conclusion.
So the limit equals \[ \boxed{0} \]