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November 7th, 2017, 01:10 AM   #1
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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:
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 February 20th, 2018, 11:10 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,505 Thanks: 1741 Sorry about delay Thanks from matheagl365
 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: 7,355 Thanks: 2469 Math Focus: Mainly analysis and algebra I would say that the limit is $e^y$. Thanks from matheagl365
 February 27th, 2018, 05:08 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,505 Thanks: 1741 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. Thanks from topsquark
February 27th, 2018, 09:36 PM   #6
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Quote:
 Originally Posted by v8archie I would say that the limit is $e^y$.
wait, you're right! but how??
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 February 27th, 2018, 10:25 PM #7 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,879 Thanks: 760 Math Focus: Wibbly wobbly timey-wimey 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: 2,090 Thanks: 1086 let's be careful here $\lim \limits_{n\to \infty}~\left(1+\left(\frac y n\right)^2\right)^{\frac n 2} = 1$ $\lim \limits_{n\to \infty}~\left(\left(1+\frac y n\right)^2\right)^{\frac n 2} = e^y$ Thanks from topsquark, v8archie and matheagl365
 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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