IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - which one gives a real asymptote as x tends to infinity

rajivranjanverma

given f(x)=(x²+3x+7)/(x+2)

to find asymptote. now if we write f(x) in quotient +remainder form we have

f(x)=(x+1)+(5/x+2).now as x becomes large f(x) approaches x+1 therefore x+1 is the asymptote.
But now if we divide numerator and denominator by x we get

f(x)=(x+3+7/x)/(1+2/x). now if x become large f(x) tends to x+3 therefore this is the asymptote
which one is the real asymptote and why? what is wrong in 2nd approach.

AAKRITI

for finding asymptote to any curve use the following conditions

(1) lim x---->infinity , dy/dx is finite say m

(2) lim x----->infinity, y - x.dy/dx is finite say k

then equation of asymptote is Y=mX+k

the answer therefore is y=x+1

rajivranjanverma

hi it was nice . but i know the ans but what i need is an explanation for my question above.why the graph is behaving in two different ways for two small alteration when x is becoming very large

rajivranjanverma

and thank u very much

meghaguptaa

Akriti can pl u prove what u have written ?

AAKRITI

megha, see.. its not a big thing,

we call a line an asymptote iff the perpendicular distance of a pt say (x,y) on the curve from the line tends to 0 when x,y tend to infinity.

we know the tangent at (x,y) is Y-y=dy/dx(X-x)

so Y=(dy/dx)X + ( y-xdy/dx )

so if asymptotes exist and x and y tend infinity, then dy/dx and y - xdy/dx should be finite limits , that's what i said before

sudeep.kumar

Rajiv,
The second aproach is incorrect, because asymptote is not simply the limit of the function when x tends to zero... this is a shortcut approach which can be used only when you express the given rational function in the form of quotient and reminder, and reminder tends to zero.
Actually the asymptote is a tangent at infinity... so if in the given implicit or explicit function of x and y, you replace y by the given asymptote and take limit x tends to infinity, you get the limit as zero. (that is why we can simply get the quotien as asymptote)

Actually the actual method is not in the syllabus, (although Aakriti has written it all correct here..)

but this shortcut may help you sometimes..
in any algebraic curve, written as a function of x and y, put the highest degree terms equal to zero to get m, the slope of asymptote.
now, limit x tends to infinity y - mx gives c. then , y = mx + c is the asymptote..

for example, in the given problem,

y(x + 2) = x² + 3x + 7, heighest degree term is x² - xy or x(x - y) = 0, so we get the slope m = 1 from x-y = 0
(there is one more asymptote corresponding to x = 0, and is parallel to y axis)

rajivranjanverma

thank u all. it was a big doubt i had. thanks a ton.

AAKRITI

you know i got this method from my test comprehension, i've always found it useful

Ask & Discuss Questions with Community & Experts