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Chapter 7: Problem 7

Evaluate \(\lim _{n \rightarrow \infty} n\left(e^{1 / n}-1\right)\).

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: Applying L'Hopital's Rule to the expression \(\frac{e^\frac{1}{n} - 1}{\frac{1}{n}}\), simplifying, and calculating the limit as n approaches infinity, we find that \(\lim_{n\to\infty} n\left(e^{1/n}-1\right) = 1\).

Step by step solution

01

Rewrite the expression

We can rewrite the expression as: \(n\left(e^{\frac{1}{n}} - 1\right) = \frac{e^\frac{1}{n} - 1}{\frac{1}{n}}\) By doing this, we have a situation with \(\frac{0}{0}\) when \(n\) approaches infinity, which is an indeterminate form.
02

Apply L'Hopital's Rule

We have the indeterminate form \(\frac{0}{0}\) as n approaches infinity. To solve this, we will apply L'H么pital's Rule, which states that \(\lim \frac{f(n)}{g(n)} = \lim \frac{f'(n)}{g'(n)}\) if both limits exist where \(f'(n)\) and \(g'(n)\) are the derivatives with respect to n. Taking the derivative of the numerator and denominator: \(f(n) = e^\frac{1}{n} - 1 \Rightarrow f'(n) = \frac{-e^\frac{1}{n}}{n^2}\) \(g(n) = \frac{1}{n} \Rightarrow g'(n) = \frac{-1}{n^2}\)
03

Apply L'Hopital's Rule

Now, applying L'H么pital's Rule, we have: \[\lim_{n\to\infty}\frac{e^\frac{1}{n}-1}{\frac{1}{n}} = \lim_{n\to\infty}\frac{\left(\frac{-e^\frac{1}{n}}{n^2}\right)}{\left(\frac{-1}{n^2}\right)}\]
04

Simplify the expression

Cancelling the common factors and simplifying, we get: \[\lim_{n\to\infty}\frac{\left(\frac{-e^\frac{1}{n}}{n^2}\right)}{\left(\frac{-1}{n^2}\right)} = \lim_{n\to\infty}e^\frac{1}{n}\]
05

Calculate the limit

As n approaches infinity, the exponent in the expression approaches 0. Therefore, we have: \[\lim_{n\to\infty}e^\frac{1}{n} = e^0 = 1\] Thus, the final answer is: \[\lim_{n\to\infty}n\left(e^{1/n}-1\right) = 1\]

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