Evaluate the following limit:
\[ \lim_{x \to 0} \frac{\ln((x^{2}+1)\cos x)}{x^{2}} = \]
______.
Answer
0.5
Explanation
The logarithm of a product splits into a sum:
\[ \frac{\ln((x^{2}+1)\cos x)}{x^{2}} = \frac{\ln(1 + x^{2})}{x^{2}} + \frac{\ln(\cos x)}{x^{2}} \]
- Since \(\ln(1 + t) \approx t\) for small \(t\), the first term tends to 1.
- Since \(\cos x \approx 1 - \frac{x^{2}}{2}\), we get \(\ln(\cos x) \approx -\frac{x^{2}}{2}\), so the second term tends to \(-\frac{1}{2}\).
The limit is \(1 - \frac{1}{2} = 0.5\). L'Hôpital's rule agrees: differentiating the top and the bottom gives \(\frac{2x/(1 + x^{2}) - \tan x}{2x}\), which also tends to \(1 - \frac{1}{2}\).
Answer: 0.5