Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 40E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 40E
Chapter:
Problem:
Prove that lim…
Step-by-Step Solution
Given information
We have to prove the following Limit\[\mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} = 0\]
Step 1: Squeeze Theorem
We will use the squeeze theorem to prove this. We know that for any sine function, \[ - 1 \le \sin \left( {\dfrac{\pi }{x}} \right) \le 1\]Take exponential on both sides:\[{e^{ - 1}} \le {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le {e^1}\]Multiply by $\sqrt x $\[\sqrt x {e^{ - 1}} \le \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le \sqrt x {e^1}\]Apply Limit on Both sides\[\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{ - 1}} \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^1}\\\sqrt 0 {e^{ - 1}} \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt 0 {e^1}\\0 \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le 0\end{array}\]So, by using the squeeze theorem, we haveTherefore, \[\mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} = 0\]
We have to prove the following Limit\[\mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} = 0\]
Step 1: Squeeze Theorem
We will use the squeeze theorem to prove this. We know that for any sine function, \[ - 1 \le \sin \left( {\dfrac{\pi }{x}} \right) \le 1\]Take exponential on both sides:\[{e^{ - 1}} \le {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le {e^1}\]Multiply by $\sqrt x $\[\sqrt x {e^{ - 1}} \le \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le \sqrt x {e^1}\]Apply Limit on Both sides\[\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{ - 1}} \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^1}\\\sqrt 0 {e^{ - 1}} \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt 0 {e^1}\\0 \le \mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} \le 0\end{array}\]So, by using the squeeze theorem, we haveTherefore, \[\mathop {\lim }\limits_{x \to {0^ + }} \sqrt x {e^{\sin \left( {\dfrac{\pi }{x}} \right)}} = 0\]