What are the horizontal Asymptotes?

Question

What are the horizontal asymptotes for the following equations:

arctan(x^(4)-x^(2))

cot^(-1) (x^(2)-x^(4))

e^(1/x)

Help would be much appreciated :)

Answer 1

If this is a homework question from beginning calculus, take the limit as x→inf.

lim (x^4 - x^2) approaches +inf
lim (x^2 - x^4) approaches -inf

These are also true when x approaches -inf as both terms are raised to even powers.

Either way, arctan( box ) as box approaches inf gives pi/2 (think about where tan goes to infinity); arccot( box) as box approaches -inf gives 0.

As far as e^(1/x) goes, 1/x goes to 0 as x gets big, so e^(1/x) goes to e^0=1 as x approaches infinity.

Answer 2

1. As x --> +/- infinity, the dominant term in arctan(x^(4)-x^(2)) is arctan(x^(4)) --> pi/2. Since the argument is squared it's the same at +/- infinity, so that there are two horizontal asymptotes with values pi/2, as x --> +/- infinity.

2. Related to that, arccot (x^(2)-x^(4)) or cot^(-1) (x^(2)-x^(4)) has horizontal asymptotes of cot^(-1) (- x^(4)) as x --> +/- infinity, but cot^(-1) ("infinity") = arctan ("1 / infinity" = 0), that is 0, so that the asymptotes have values zero (0) as x --> +/- infinity.

3. e^(1/x) does NOT have a horizontal asymptote as x--> 0 it has VERTICAL asymptotes. (1/x and therefore e^(1/x) is UNBOUNDED as x --> 0.)

However, 1/x --> 0 as x --> +/- infinity, and thus e^(1/x) has asymptotes with values of e^0 (that is, 1) as x --> +/- infinity.

Live long and prosper.

Answer 3

vertical asymptote is the place y is undefined enable y = f(x)= -2x+2/x subsequently y = -2x^2 + 2 = 2( -x^2 -a million ) = 2 (a million+x) (a million-x) subsequently y is undefined at x = -a million and x = a million those are vertical asymptotes horizontal asymptotes is fee of y whilst x techniques infinity : y = -2x^2 + 2 whilst x is going to infinity +2 and coffecient of x^2 will become negligible subsequently y= -x^2 that's y techniques adverse infinity : )

Answer 4

Usually, horizontal asymptotes are associated with hyperbolic functions, and indicate the range of y for the domain of x. The interval BETWEEN the asymptotes is not allowed for y.

For the arc-tan and the arc-cot (the second one), the horizontal asymptotes occur every interval between -pi/2 and pi/2, and +/- pi above and below that interval. With the arc-tan function, the range for graphing needed is from -pi/2 to positive infinity; with the arc-cot function, the range is from pi/2 as a maximum to negative infinity.

Answer 5

e^(1/x) has a horizontal asymptote at zero