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 December 7th, 2006, 07:53 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?
 December 7th, 2006, 10:20 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)?
 December 8th, 2006, 01:23 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
 December 8th, 2006, 05:13 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
December 9th, 2006, 05:28 AM   #5
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Quote:
 Originally Posted by Mazaheri 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.
OK. You have a line with a (nonvertical) linear asymptote as x increases without bound. Knowing this, you're right that m = lim_{x-->infinity} f(x)/x.

Quote:
 Originally Posted by Mazaheri So let's say lim f'(x)= lim (f(x)/x)when x goes to infinit.
Also correct, under the assumptions above (nonvertical linear asymptote as x "approaches infinity" = increase without bound).

Quote:
 Originally Posted by Mazaheri 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
I don't think that this method works in general. Are you thinking of L'Hopital's Rule?

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