Im trying to solve this problem using l'hopital but amm not sure how to do it
lim
X->infinite x^3 * e^(-x^2)
soo this infinite * e^-infinite... but from there I am not sure if you can use it to solve this...
ok well after asking arround i got to the answer... thanks for the help but the thing is I am supposed to just use the properties to reach other side... i can't alter it... but i guess is my fault for not specifying... well the way it went was quite tricky and required a property i don't use...
more trig...
Amm I've been at this problem for an hour already it loooked really easy but for some reason i can't reach the answer
(tan x + sec x)^2 = (1 + sin x)/(1 - sin x)
soo what went and tried was expand it and then exchage tan and sec..
(sin^2 x /cos^2 x )+ (2/cos x) + (1/ sin^2 x)...
wait ... where did that
2\pm\sqrt{10}
edit: i found i see where you got it but still... I am lost..
the original range for x was ]0,4[...
and one more thing I am seeing here that there is a table in which you put this values to see if they are positive or negative once evaluated... so...
so if that is that and i have a restriction on the lower value i can multiply without reversing the sign right?? soo
x^2 + 3x + 1 = 7(x+1)
X^2 - 4x -6 = 0
and then use -b formula thing
a = 1, b = -4, c = -6
(4+-sqrt(4^2 - 4(1)(-6)))/2
(4+- sqrt (40))/2
? for some reason it sound like I am...
ok so i missed a few classes and i have a test comming up soo and i need some help understanding inequalities.
the example I am trying to do is:
If |x-2|< 2, show that |(x^2+3x+1)/(x+1)| < 7
form what i could understand from a textbook was something like:
-2<x-2<2
0 < x < 4
and...