Posted by Anonymous on .
What is the horizontal asymptote of f(x)=x/(x1)^2
It is supposed to be y=0 but then I even know that b/d is the horizontal asymptote, in this since there is no b, it would be 0/1 which is 0
is the above reasoning correct?

Math  horizontal asymptote of f(x) (check) 
jai,
horizontal asymptote occurs where the value of x is restricted,,
n the given function f(x)=x/(x1)^2, what value of x is restricted? 
Math  horizontal asymptote of f(x) (check) 
jai,
oops i got it wrong
horizontal asymptote occurs where the value of y is restricted,,
so in the given function f(x)= y = x/(x1)^2, what value of y is restricted? 
Math  horizontal asymptote of f(x) (check) 
Reiny,
your function expanded is
f(x) = x/(x^2  2x + 1)
For a horizontal asymptote we look at what happens to the function as x > infinity.
Use an intuitive approach ...
as x becomes very large, say x = 1 million, the denominator becomes large much faster than the numerator.
So you have a division by a hugely large number resulting in a number close to zero
So when x > + infinity, f(x) > +0 (still above the xaxis)
wen x >  infinity , f(x) > 0 (slightly below the xaxis) 
Math  horizontal asymptote of f(x) (check) 
Anonymous,
is it that the yvalue is restricted to being positive. sine the horizontal asymptote has an equation of y=0.

Math  horizontal asymptote of f(x) (check) 
jai,
horizontal asymptote:
y = 0 , as x approaches plus/minus infinity