\( \lim _{n \rightarrow \infty } \frac{n\cdot \sin \frac{1}{n}-\cos \frac{1}{n}}{\ln (1+\frac{1}{n^{2}})} \)

To solve the limit \[ \lim _{n \rightarrow \infty } \frac{n \cdot \sin \frac{1}{n}-\cos \frac{1}{n}}{\ln (1+\frac{1}{n^{2}})}, \] we will analyze the numerator and the denominator separately as \( n \) approaches infinity. ### Step 1: Simplifying the numerator First, we will evaluate the expression in the numerator as \( n \) approaches infinity: \[ n \cdot \sin \frac{1}{n} - \cos \frac{1}{n}. \] Using the small angle approximation for sine, we have \( \sin x \approx x \) when \( x \) is small. Thus, \[ \sin \frac{1}{n} \approx \frac{1}{n} \Rightarrow n \cdot \sin \frac{1}{n} \approx n \cdot \frac{1}{n} = 1. \] Next, we can expand \( \cos \frac{1}{n} \) using the Taylor series expansion: \[ \cos x \approx 1 - \frac{x^2}{2} \text{ for small } x. \] Therefore, \[ \cos \frac{1}{n} \approx 1 - \frac{(1/n)^2}{2} = 1 - \frac{1}{2n^2}. \] Putting this back into our expression for the numerator: \[ n \cdot \sin \frac{1}{n} - \cos \frac{1}{n} \approx 1 - \left( 1 - \frac{1}{2n^2} \right) = \frac{1}{2n^2}. \] ### Step 2: Simplifying the denominator Now, we analyze the denominator: \[ \ln \left( 1 + \frac{1}{n^2} \right). \] For small \( x \), the Taylor series expansion for \( \ln(1+x) \) gives \( \ln(1+x) \approx x \). Hence, \[ \ln \left( 1 + \frac{1}{n^2} \right) \approx \frac{1}{n^2}. \] ### Step 3: Combining the results Now we can substitute the simplified numerator and denominator back into the limit: \[ \lim_{n \rightarrow \infty} \frac{n \cdot \sin \frac{1}{n} - \cos \frac{1}{n}}{\ln \left( 1 + \frac{1}{n^2} \right)} \approx \lim_{n \rightarrow \infty} \frac{\frac{1}{2n^2}}{\frac{1}{n^2}}. \] This simplifies to: \[ \lim_{n \rightarrow \infty} \frac{1/2}{1} = \frac{1}{2}. \] ### Conclusion Thus, the limit is \[ \boxed{\frac{1}{2}}. \]