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  L'Hospital rule.......a perfect outlook.   15 Nickels awarded!
Tagged with:    [Post New]posted on 18 Apr 2007 19:38:56 IST    
Let lim stand for the limit lim_(x->c), lim_(x->c^-), lim_(x->c^+), lim_(x->infty), or lim_(x->-infty), and suppose that lim f(x) and lim g(x) are both zero or are both +/-infty. If
lim(f^'(x))/(g^'(x))
(1)
has a finite value or if the limit is +/-infty, then
lim(f(x))/(g(x))==lim(f^'(x))/(g^'(x)).
L'Hospital's rule occasionally fails to yield useful results, as in the case of the function lim_(u->infty)u(u^2+1)^(-1/2). Repeatedly applying the rule in this case gives expressions which oscillate and never converge,
lim_(u->infty)u/((u^2+1)^(1/2)) = lim_(u->infty)1/(u(u^2+1)^(-1/2))
(3)
= lim_(u->infty)((u^2+1)^(1/2))/u
(4)
= lim_(u->infty)(u(u^2+1)^(-1/2))/1
(5)
= lim_(u->infty)u/((u^2+1)^(1/2)).
(6)
The actual limit is 1.
LHospitalsRule1
L'Hospital's rule must sometimes be applied with some care, since it holds only in the implicitly understood case that g^'(x) does not change sign infinitely often in a neighborhood of infty. For example, consider the limit f(x)/g(x) with
f(x) = x+cosxsinx
(7)
g(x) = e^(sinx)(x+cosxsinx)
(8)
as x->infty. While both f(x) and g(x) approach infty as x->infty, the limit of the ratio is bounded inside the interval [1/e,e], while the limit of f^'(x)/g^'(x) approaches 0 (Boas 1986).
LHospitalsRule2
Another similar example is the limit f(x)/g(x) with
f(x) = xsin(x^(-4))e^(-1/x^2)
(9)
g(x) = e^(-1/x^2)
(10)
as x->0. While both f(x) and g(x) approach 0 as x->0, the limit of the ratio is 0, while the limit f^'(x)/g^'(x) is unbounded on the real line
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kghedriu (2333)

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