Question

If lim x-> infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b

Answer 1

We have

`(x^2)/(x+1)=((x+1)^2-2(x+1)+1)/(x+1)=(x+1)-2+\frac1{x+1}=x-1+\frac1{x+1}`

So

`\displaystyle\lim_(x\to\infty)\left((x^2)/(x+1)-ax-b\right)=\lim_(x\to\infty)\left(x-1+\frac1{x+1}-ax-b\right)=0`

The rational term vanishes as x gets arbitrarily large, so we can ignore that term, leaving us with

`\displaystyle\lim_(x\to\infty)\left((1-a)x-(1+b)\right)=0`

and this happens if a = 1 and b = -1.

To confirm, we have

`\displaystyle\lim_(x\to\infty)\left((x^2)/(x+1)-x+1\right)=\lim_(x\to\infty)(x^2-(x-1)(x+1))/(x+1)=\lim_(x\to\infty)\frac1{x+1}=0`

as required.