I'm taking a massively blank stab at this one for my procrastination from accounting, so the quality of this answer will be again, poor lol.
Wouldn't you have to also check/assume the individual requirements of L'Hs are satisfied, namely the limits have to all exist?
Change variables.
P(e^{x})/Q(x)
Without Loss of Generality, we can assume P is of degree 1.
This expression obviously tends to infinity.
For those who do not see it, apply L'Hôpital's Rule (with respect to x) n times where n is the degree of Q.
If P is larger, then the contribution to the explosion is greater, and the expression still diverges.
[QED]
Last edited by Paradoxica; 13 Jun 2016 at 1:43 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Yes. I would, and I can, and I will. I am quasi-mentoring (mainly because he is very self capable) a year 10 student who has self-taught all the basics of Extension 2 Integration, and I will be teaching him competitive integration. Then I will drag him to the MATHSOC Integration Bee next year.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Going to infinity counts as "exist" for the purposes of using L'Hôpital's rule. You can't have something that oscillates forever though (like (2+cos(x))/(2-cos(x)) or something), that'd fail to satisfy the requirement for L'Hôpital's rule.
Prove that f is differentiable at the origin.
There is a real positive number θ which satisfies tan^{-1}θ = sinθ.
Prove that f is continuous at θ.
Hence prove f is continuous at -θ.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Yep. MVT implies (f(x)-f(a))/(x-a)=f'(c(x)) for some c(x) strictly between a and x. As x->a, c(x)->a, which implies that the RHS converges to a limit by the question's assumptions. So the LHS converges to a limit as x->a, which is precisely the definition of differentiability.
A related followup:
Let f:R->R be a differentiable function. Must f' be continuous?
Last edited by seanieg89; 2 Jul 2016 at 10:28 AM.
And another related Q.
Last edited by InteGrand; 2 Jul 2016 at 3:22 PM.
Bingo. Haha I figured you would realise that the previous exponent wasn't good enough when you tried to work out the details.
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