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Chapter 1: Problem 63

\(\lim _{x \rightarrow \infty} \sqrt[x]{2 \sum_{n=0}^{x} \frac{x^{n}}{n !}}\) is equal to (A) (B) \(\mathrm{e}\) (C) \(2 \mathrm{e}^{-1}\) (D) 0

Short Answer

Expert verified
Question: Evaluate the limit - \(\lim _{x \rightarrow \infty} \sqrt[x]{2 \sum_{n=0}^{x} \frac{x^{n}}{n !}}\). Answer: (B) \(\mathrm{e}\).

Step by step solution

01

Recall the Taylor series expansion of the exponential function

The Taylor series expansion for \(e^{x}\) is given by \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\). We will use this to help evaluate the given limit.
02

Rewrite the summand using the Taylor series

Divide the summand by \(x^n\) on each side: \(\frac{2 \sum_{n=0}^{x} \frac{x^{n}}{n!}}{x^x}=\frac{2}{x^x}\sum_{n=0}^x \frac{x^n}{n!}\).
03

Use the relationship between the Taylor series and the exponential function

Since we have the sum up to x terms, consider the remaining infinite terms as y and get the complete expression for e^x: \(\frac{2}{x^x}\sum_{n=0}^x \frac{x^n}{n!} \implies \frac{2}{x^x}(e^x - y)\).
04

Rewrite the expression to calculate the limit

In order to find the limit, rewrite the expression as: \(\sqrt[x]{2(e^x - y)}\).
05

Apply the limit to the expression

Now, evaluate the limit: \(\lim_{x\rightarrow\infty}\sqrt[x]{2(e^x - y)}\). As x tends to infinity, the term y becomes negligible, therefore: \(\lim_{x\rightarrow\infty}\sqrt[x]{2(e^x - y)} \approx \lim_{x\rightarrow\infty}\sqrt[x]{2e^x}\).
06

Simplify the expression and evaluate the limit

Rewrite the expression: \(\lim_{x\rightarrow\infty}\sqrt[x]{2e^x}=\lim_{x\rightarrow\infty}(2e^x)^{\frac{1}{x}}\). Since the exponent is a fraction, we can write the expression as \(\lim_{x\rightarrow\infty}2^{1/x}e\), and the limit becomes: \(\lim_{x\rightarrow\infty}2^{1/x}e\). As x approaches infinity, \(2^{1/x}\) approaches 1, and therefore: \(\lim_{x\rightarrow\infty}2^{1/x}e = 1 \cdot e=e\). The correct answer is (B) \(\mathrm{e}\).

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