Date: 23 Apr 1995 01:24:03 -0400 From: Naseema Nabi Subject: HELP! Dear Dr. Math, Help me, I can't figure out how to take the limit using L'Hopital's rule on this one problem. Lim [(arctan X - (3.14(Pie)/4)/(X-1)] x=1 Please contact me soon at firstname.lastname@example.org
Date: 23 Apr 1995 12:49:56 -0400 From: Dr. Sydney Subject: Re: HELP! Dear Naseema, Hello! I think on this problem you actually don't need to use L'Hopital's Rule. However, I am a bit unclear on what the problem is. I think the way you have written your parentheses the problem is this: lim arctan x - pi/4(x-1) x->1 (So, the x-1 is in the denominator) In this case, think about what happens to the last term (the pi/4(x-1) term) as x approaches one. The closer you get to one, the smaller (x-1) becomes, and the bigger pi/4(x-1) becomes, right? (Note: this is in the case that you are approaching one from numbers bigger than one. Can you see what happens if you approach one by numbers smaller than one?). So, there is no limit as x approaches one. If, however, the problem is lim arctan x - pi(x-1)/4 x->1 Then, split the limit up into the sum of two limits, and so you have lim arctan x - lim pi(x-1)/4 x->1 x->1 Evaluate these two limits (you still won't need to L'hop...) and you'll have your answer! If any of this is unclear, or if you have any other questions feel free to write back! --Sydney, "Dr. Math!"
Date: 23 Apr 1995 15:04:43 -0400 From: Dr. Ken Subject: Re: HELP! Hello there! I'm another math doctor here, and I interpreted your message in yet another way. It looks to me like you have Arctan(x) - Pi/4 Lim x->1 ------------------ x-1 Is this right? Then you _would_ use L'Hopital's rule; let's review the criteria for when you can use L'Hopital. Use it when the numerator and the denominator are both going to positive infinity, or when they're both going to zero, or when they're both going to negative infinity. In this case, if you plug in 1 for x in the to and the bottom, you'll get zero. So Yippee, we can use L'Hopital! That means we can take the derivative of the top and bottom. The derivative of the top is 1/(1 + x^2), and the derivative of the bottom is 1. So we have 1 1 1 Lim x->1 --------- = ------- = - . 1 + x^2 1 + 1 2 I hope this helps! -Ken Dr. Math
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