AboutAbe Mantell Expertise Hello,
I am a college professor of mathematics and regularly teach all levels
from elementary mathematics through differential equations, and would
be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
Hi Mantell, i really wana solve the questions qhich i m sending you. i actully already solve it but need not reconfirm it.
Thanks
Attir
Answer OK...
a) Divide each term by x to get: lim(x->oo) (1+2cos(x)/x)/(4+sin(x)/x)
.. so the cos(x)/x and sin(x)/x both approach zero as x->oo, so the
.. limit is 1/4.
b) Also dividing by x (in the numerator and denominator), we get
.. lim(x->oo) (4+3/x)/sqrt(1+2/x), since the "/x" goes in the "sqrt"
.. as "/x^2"...the 3/x and 2/x terms approach zero...so the limit is
.. 4/sqrt(1)=4/1 = 4.
c) First divide each term by x, to get: lim(x->oo) [(1+a/x)/(1-a/x)]^x
.. which equals: lim (1+a/x)^x / lim (1-a/x)^x. You may have already
.. done in class, or the text book showed, using L'Hopital's Rule that
.. lim(x->oo) (1+a/x)^x = e^a. So, we get e^a/e^(-a) = e^(2a).
.. If we wan that to equal e^3, then a=3/2.
)