Mazaheri

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?

CRGreathouse

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)?

Mazaheri

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

Cat

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

CRGreathouse

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.

Also correct, under the assumptions above (nonvertical linear asymptote as x "approaches infinity" = increase without bound).

I don't think that this method works in general. Are you thinking of L'Hopital's Rule?