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December 7th, 2006, 08:53 AM   #1
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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
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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
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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
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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
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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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