#determine a b c
ruby trail
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ruby trail
yes
ruby trail
no
gaunt veldt
L hospital is a very very bad idea
ruby trail
need to just rationalise it
gaunt veldt
it wont even work here lol
brisk flame
ruby trail
so ax + the sqrt
ruby trail
yes
gaunt veldt
also try taking the x out from the sqrt
jade lark
multiply by 1 to lose the sq root in the numerator
gaunt veldt
square both sides?
jade lark
$$ \frac{(a^2-c)x^2 -bx +2}{a + \sqrt{c+\frac{b}{x} - \frac{2}{x^2}}} \xrightarrow[x\to\infty]{} 1 $$
fast condor
ca n someone please write it down on a paper cause i dont really understand
they are just asking you rationalize the terms , to get the sqrt out of this , the denominator then become determinate [ aka you can get that shit out of the limit] . then you take stuff common from the brackets
tulip abyssBOT
aL
fast condor
they are just asking you rationalize the terms , to get the sqrt out of this , t...
point of taking the common terms is that 1/x is 1/infinity is zero , which helps us
then you get it into a standard form -> thats probably your required condition
or else you solve the eqn you get with additional info
jade lark
since x gets large you must take a^2 - c = 0 = b
ruby trail
yea that s true
gaunt veldt
x(ac-sqrt(cx^2+bx-2))
=x(a^2x^2-cx^2-bx+2)/(ax+sqrt(cx^2+bx-2))
=x((a^2-c)x^2-bx+2)(x(a+sqrt(c+b/x-2/x^2))
=((a^2-c)x^2-bx+2)/(a+sqrt(c+b/x-2/x^2))
~((a^2-c)x^2-bx+2)/(a+sqrt(c))
clearly it blows up unless x^2 and x have zero coeffecients
a^2-c=-b=0
2/(a+sqrt(c))=1
might be erroneous im typing without glasses
there i missed a c