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

Question

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To evaluate the limit:

$$
\lim_{n \rightarrow \infty} \frac{n \cdot \sin \frac{1}{n} - \cos \frac{1}{n}}{\ln \left(1 + \frac{1}{n^{2}}\right)}
$$

we can perform a substitution to simplify the expression. Let $$ x = \frac{1}{n} $$, so as $$ n \to \infty $$, $$ x \to 0 $$. Rewriting the limit in terms of $$ x $$:

$$
\lim_{x \rightarrow 0} \frac{\frac{\sin x}{x} - \cos x}{\ln(1 + x^{2})}
$$

Next, expand the trigonometric functions and the logarithm using their Taylor series around $$ x = 0 $$:

1. **Sine and Cosine Expansions:**
   $$
   \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots
   $$
   $$
   \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots
   $$
   Therefore,
   $$
   \frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \dots
   $$
   
   Subtracting the cosine term:
   $$
   \frac{\sin x}{x} - \cos x = \left(1 - \frac{x^2}{6} + \frac{x^4}{120} - \dots\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots\right) = \frac{x^2}{3} - \frac{x^4}{30} + \dots
   $$

2. **Logarithm Expansion:**
   $$
   \ln(1 + x^{2}) = x^{2} - \frac{x^4}{2} + \frac{x^6}{3} - \dots
   $$

Now, the limit becomes:

$$
\lim_{x \rightarrow 0} \frac{\frac{x^2}{3} - \frac{x^4}{30} + \dots}{x^{2} - \frac{x^4}{2} + \dots} = \lim_{x \rightarrow 0} \frac{\frac{1}{3} - \frac{x^2}{30} + \dots}{1 - \frac{x^2}{2} + \dots}
$$

As $$ x \to 0 $$, higher-order terms ($$ x^2 $$, $$ x^4 $$, etc.) become negligible, leaving:

$$
\frac{1/3}{1} = \frac{1}{3}
$$

**Final Answer:**

$$\boxed{\dfrac{1}{3}}$$
The limit is $$ \frac{1}{3} $$.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Step 1: Simplifying the numerator

Step 2: Simplifying the denominator

Step 3: Combining the results

Conclusion

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