91Ó°ÊÓ

We start by rewriting the function in terms of a quotient (division), so that we can apply L'Hopital's Rule. Given function: \( \lim _{n \rightarrow \infty} n \tan \frac{1}{n} \) Rewritten as: \( \lim _{n \rightarrow \infty} \frac{\tan \frac{1}{n}}{\frac{1}{n}} \)

Now that we have rewritten the function, let's apply L'Hopital's Rule. This rule states that if the limit \( \lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} \) results in an indeterminate form (0/0 or ∞/∞), then we can evaluate the limit by taking the derivatives of both f(n) and g(n) and finding the new limit: \( \lim _{n \rightarrow \infty} \frac{f'(n)}{g'(n)} \) Now, find the derivatives of both the numerator and denominator: \( f(n) = \tan \frac{1}{n} \) => \( f'(n) = \frac{-1}{n^2} \sec^2 \frac{1}{n} \) \( g(n) = \frac{1}{n} \) => \( g'(n) = -\frac{1}{n^2} \)

Now, we plug the derivatives back into the limit and compute the result: \( \lim _{n \rightarrow \infty} \frac{\frac{-1}{n^2} \sec^2 \frac{1}{n}}{-\frac{1}{n^2}} \) As we can see, the \( -\frac{1}{n^2} \) in both numerator and denominator cancel out: \( \lim _{n \rightarrow \infty} \sec^2 \frac{1}{n} \) Now, we have a much simpler function to deal with. Since the secant function is reciprocal of the cosine function, rewrite the function in terms of cosine: \( \lim _{n \rightarrow \infty} \frac{1}{\cos^2 \frac{1}{n}} \) As n approaches infinity, \( \frac{1}{n} \) approaches 0, so we compute the limit: \( \lim _{n \rightarrow \infty} \frac{1}{\cos^2 \frac{1}{n}} = \frac{1}{\cos^2 0} \) Since \( \cos(0) = 1 \), the limit evaluates to: \( \frac{1}{1^2} = 1 \) Therefore, \( \lim _{n \rightarrow \infty} n \tan \frac{1}{n} = 1 \).