November 7th, 2017, 01:10 AM  #1 
Newbie Joined: Nov 2017 From: Europe Posts: 2 Thanks: 0  How to calculate this limit?
Hi! I know that the value of this limit is 1, but I do not know how to calculate it:

February 20th, 2018, 11:10 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,045 Thanks: 1618 
Sorry about delay

February 27th, 2018, 01:45 PM  #3 
Newbie Joined: Feb 2018 From: United States Posts: 6 Thanks: 0 Math Focus: Calculus, Linear Algebra 
Hi, First, as n approaches infinity, the term (y/n)^2 approaches 1; hence, the term (1+(y/n)^2) approaches 1. Even though n/2 will approach infinity, 1 to any power equals 1. Hope that helps! 
February 27th, 2018, 03:21 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
I would say that the limit is $e^y$.

February 27th, 2018, 05:08 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,045 Thanks: 1618 
The limit is 1, but matheagl365's argument is flawed. Consider the natural logarithm of the expression. It's easy to show that its limit is zero. 
February 27th, 2018, 09:36 PM  #6 
Newbie Joined: Feb 2018 From: United States Posts: 6 Thanks: 0 Math Focus: Calculus, Linear Algebra  wait, you're right! but how??

February 27th, 2018, 10:25 PM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff. 
skipjack has the correct answer. You can get the expression by taking the natural log of both sides of L = lim(...). Dan 
February 27th, 2018, 10:33 PM  #8 
Senior Member Joined: Sep 2015 From: USA Posts: 1,974 Thanks: 1025 
let's be careful here $\lim \limits_{n\to \infty}~\left(1+\left(\frac y n\right)^2\right)^{\frac n 2} = 1$ $\lim \limits_{n\to \infty}~\left(\left(1+\frac y n\right)^2\right)^{\frac n 2} = e^y$ 
March 1st, 2018, 02:10 PM  #9 
Newbie Joined: Feb 2018 From: United States Posts: 6 Thanks: 0 Math Focus: Calculus, Linear Algebra 
oh, thanks for pointing that out.


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