Question

In Exercises \(27-30,\) find the limit of \(s(n)\) as \(n \rightarrow \infty\) $$ s(n)=\frac{18}{n^{2}}\left[\frac{n(n+1)}{2}\right] $$

Short answer

The limit of the function \(s(n)\) as \(n\) tends to infinity is 9.

Step-by-step solution

Simplify the algebraic expression

We start by simplifying the expression for \(s(n)\). This can be done by multiplying through the fraction present, which will result in: \ns(n) = \( \frac{18n(n+1)}{2n^2} \). Simplifying this, further gives: \(s(n) = \frac{9(n+1)}{n}\)

Apply the limit definition

In this step, we will set the limit of \(s(n)\) as \(n\) tends towards infinity, which translates to: \(\lim_{n \rightarrow \infty} s(n) = \lim_{n \rightarrow \infty} \frac{9(n+1)}{n}\). This will leave us with \( \frac{9 \cdot \infty}{\infty} \). Dividing infinity by infinity is called an indeterminate form, and we cannot directly say what the limit is from this form.

Applying L'Hôpital's Rule

L'Hôpital's Rule is a way to evaluate these sorts of indeterminate forms. It states that when trying to evaluate \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) where the limit is of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the limit will also be \( \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \) which is \( \frac{\text{d}}{\text{d}x} \left( \frac{f(x)}{g(x)} \right) \). On applying this to our limit, we differentiate \( 9(n+1) \) and \( n \) to obtain \(9\) and \(1\) respectively. The limit then simplifies to \( \frac{9}{1} \).