Problem 22 - Calculus Early Transcendentals

Question

Evaluate the following limits using Taylor series. $$\lim _{x \rightarrow \infty} x\left(e^{1 / x}-1\right)$$

Step-by-Step Solution

Answer

Based on the given step by step solution, evaluate the following limit using Taylor series: \(\lim_{x \rightarrow \infty} x\left(e^{1 / x}-1\right)\) Answer: The limit evaluates to \(1\).

Step 1: Find the Taylor series expansion of \(e^{1/x}\) around \(x=\infty\)

We can find the Taylor series expansion for \(e^{1/x}\) by first recognizing that the function that we want to expand is \(f(u) = e^u\) (where we will substitute \(u=1/x\)). The Taylor series expansion for \(e^u\) is given by: $$e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!} = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$$ Now, we will substitute \(u=1/x\): $$e^{1/x} = 1 + \frac{1}{x} + \frac{1}{2!x^2} + \frac{1}{3!x^3} + \cdots$$

Step 2: Substitute the Taylor series expansion into the given expression

Now, we will substitute the Taylor series expansion we found in step 1 into the given expression: $$x(e^{1/x} - 1) = x\left[\left(1 + \frac{1}{x} + \frac{1}{2!x^2} + \frac{1}{3!x^3} + \cdots\right) - 1\right]$$

Step 3: Simplify the expression and take the limit

Simplify the expression: $$x(e^{1/x} - 1) = x\left(\frac{1}{x} + \frac{1}{2!x^2} + \frac{1}{3!x^3} + \cdots\right) = 1 + \frac{1}{2!x} + \frac{1}{3!x^2} + \cdots$$ Now, we will take the limit as \(x\) approaches infinity: $$\lim_{x \rightarrow \infty} x\left(e^{1/x} - 1\right) = \lim_{x \rightarrow \infty} \left(1 + \frac{1}{2!x} + \frac{1}{3!x^2} + \cdots\right)$$ As \(x\) approaches infinity, each term in the parentheses with a factor of \(1/x\) or higher will approach zero, leaving only the first term: $$\lim_{x \rightarrow \infty} x\left(e^{1/x} - 1\right) = 1$$ So, the final result for the given limit is: $$\lim _{x \rightarrow \infty} x\left(e^{1 / x}-1\right) = 1$$