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December 7th, 2006, 08:53 AM | #1 |
Member Joined: Dec 2006 Posts: 65 Thanks: 0 | lim
I have found a manner to calculate some infinite limits but I couldn't find a precise proof for it .If we calculate the m of an asymptote line y=mx+n of a curve y=f(x) by two ways,limf'(x)=limf(x)/x when x goes to infinite that seems to be a right fact ,some limits become easier.Fpr example lim ln(x)/x= lim 1/x.Would you please help me?
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December 7th, 2006, 11:20 AM | #2 |
Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms |
If you know the equation of the asymptote, that would make finding the limit easier. I'm not entirely sure what you're saying, though. Could you give an example with a more complex function (so I can see what has a derivative taken and what doesn't, and where they go)? |
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December 8th, 2006, 02:23 AM | #3 |
Member Joined: Dec 2006 Posts: 65 Thanks: 0 |
Thank you for your reply .I will try to say my idea more precisely. Let's line y=mx+n be asymptote of curve y=f(x).Indeed in infinit this line is tangent to curve ,so no carefuly we can say m=lim f'(x) when x goes to infinit.On the other hand there is a precise proved manner for calculating m.It is m=lim (f(x)/x)when x goes to infinit. So let's say lim f'(x)= lim (f(x)/x)when x goes to infinit.For example: lim e^x/x=lim e^x=infinit when x goes to infinit lim((x^2+1)^.5-(x^2+2)^.5 )/x =lim(x/(x^2+1)^.5-x/(x^2+2)^.5)=0 when x goes to infinit. While the ordinery calculation is a little more difficault. There is more difficault examples indicating the benifie of this manner but I would like to see a precise proof for it to use it surely. Regards |
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December 8th, 2006, 06:13 PM | #4 |
Newbie Joined: Dec 2006 Posts: 29 Thanks: 0 |
I think you are referring to L'Hopital's Rule which states the following: Let f and g be functions such that lim(x->a) f/g is either infinity/infinity or 0/0, then the limit may be computed alternatively as lim(x->a) f' / g' . As in one of your examples, lim(x->infinity) e^x/x = f/g where f = e^x and g = x. We can instead take lim(x->infinity) f' / g' = lim(x->infinity) e^x/1 = infinity. So, I believe what you have is the following theorem: Let f be a function such that lim(x->infinity) f is infinity. Then, lim(x->infinity) f/x = lim(x->infinity) f' . That's a very nice observation, and I hope that this helps you. -Cat |
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December 9th, 2006, 06:28 AM | #5 | |||
Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms | Quote:
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