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Chapter 1: Q. 93 (page 151)

Prove the $k = 1$ case of the second part of Theorem $1.31$ (b): that $\lim_{x \to \infty} e^{- x} = 0$ .

Short Answer

Expert verified

It is proved for theorem $1.31 (b)$ that $\lim_{x \to \infty} e^{- x} = 0$ .

Step by step solution

Step 1. Given Information

We are given,

$\lim_{x \to \infty} e^{- x} = 0$

Step 2. Proving the statement

The limit is given by,

$\lim_{x \to \infty} e^{- x} = \lim_{x \to \infty} \frac{1}{e^{x}} = \frac{1}{e^{\infty}} = \frac{1}{\infty} = 0$

Hence Proved.

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Prove the k=1 case of the second part of Theorem 1.31(b): that limx→∞e-x=0.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the statement

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Prove the $k = 1$ case of the second part of Theorem $1.31$ (b): that $\lim_{x \to \infty} e^{- x} = 0$ .