Squeeze theorem problem

Question

Use the Squeeze theorem to prove lim(x→∞) 1/x+e^-x = 0

For x ≥ 0, since 0 < e^-x ≤ 1, we have

_____ < x + e^-x < _____

Therefore,

_______ < 1/x+e^-x < _______

please fill in the blanks and explain how you got the answer

Answer 1

A) For x ≥ 0, since 0 < e^-x ≤ 1, we have

B) __x___ < x + e^-x < __1+x___

Therefore,

C) ___1/(1+x)____ < 1/x+e^-x < ___1/x____

Explanation

For part A: The setup of the Squeeze Theorem requires that we start with what we know for sure. 0 < e^-x ≤ 1. Note that e^-x =1/(e^x). If x approaches a very big number, we can say that e^-x approaches 0 and if x approaches a very small number close to 0, we can say that e^-x approaches 1. Therefore, this setup makes sense.

For part B: From the first setup, we are going to add x to each side of the inequality. Therefore, x+0 < x+e^-x ≤ x+1.

For part C: Then we can take the reciprocal of each side 1/x < 1/(x+e^-x ) ≤1/(x+1). However, the inequality signs switch because the value of the numbers would switch when we take the reciprocal (for example, 4<5, but 1/4>1/5). Therefore, the last blank is 1/(1+x) < 1/(x+e^-x ) < 1/x.

To finish the proof, lim(x→∞) 1/(1+x) < lim(x→∞) 1/(x+e^-x ) < llm(x→∞) 1/x, which leads to 0 < lim(x→∞) 1/(x+e^-x ) <0. The only way that a number can be between 0 and 0 is if the middle is also 0. Therefore, lim(x→∞) 1/x+e^-x = 0.