Please solve this question

if lim [(x^2 + x +1)/(x+1) -ax-b]=4
x--->∞
then, find the values of a & b ..

lim [(x^2 + x +1)/(x+1) -ax-b]=4
x--->∞
=>lim [(x^2 + x +1-ax^2-bx-ax-b)/(x+1) ]=4
x--->∞
=>lim [((1-a)x^2+(1-a-b)x+(1-b))/(x+1)]=4
x--->∞
Let y=1/x
so, now our structure is
lim [((1-a)/y^2+(1-a-b)/y+(1-b))/(1/y+1)]=4
y--->0
=>lim [((1-a)+(1-a-b)y+(1-b)y^2)/(y+y^2)]=4
y--->0
Here we can use L' Hospital rule if the upper part of limit be 0 when y--->0
so we get
1-a=0 => a=1
Now applying L' Hospital rule
lim [((1-a-b)+2(1-b)y)/(1+2y)]=4
y--->0
=>1-a-b=4
=>b=-4
So, The values of a=1 and b=-4

Hello,

It's easy if you can remember the cubic identity:
a³ - b³ = (a - b)(a² + ab + b²)

Thus by having a=x and b=1:
x³ - 1 = (x - 1)(x² + x + 1)
x² + x + 1 = (x³ - 1)/(x - 1)
(x² + x + 1)/(x + 1) = (x³ - 1)/[(x - 1)(x + 1)]
(x² + x + 1)/(x + 1) = (x³ - 1)/(x² - 1)
(x² + x + 1)/(x + 1) = (x³ - x)/(x² - 1) + x/(x² - 1)
(x² + x + 1)/(x + 1) = x + x/(x² - 1)
(x² + x + 1)/(x + 1) - ax - b = (1 - a)x - b + x/(x² - 1)

So:
Lim (x→+∞) [(x² + x + 1)/(x + 1) - ax - b]
   = Lim (x→+∞) [(1 - a)x - b + x/(x² - 1)]
   = Lim (x→+∞) [(1 - a)x]    +    Lim [(x→+∞) x/(x² - 1)]    - b
   = 4

We know that:
Lim [(x→+∞) x/(x² - 1)] = 0
Lim (x→+∞) [(1 - a)x] = +∞ if a<1
Lim (x→+∞) [(1 - a)x] = -∞ if a>1
Lim (x→+∞) [(1 - a)x] = 0   if a=1

In order to have a finite limit, we must have:
Lim (x→+∞) [(1 - a)x] = 0
So a must be equal to 1.

Thus:
Lim (x→+∞) [(x² + x + 1)/(x + 1) - ax - b] = -b = 4
and b=-4.

Thus (a; b) = (1; -4)

Regards,
Dragon.Jade :-)

((x+1)^2 - x) / (x+1)

x+1 - x/(x+1)

if x approaches infinity then limit of second term is 1.

so you have

x - ax - b = 4

x(1-a)-b = 4

a = 1, b = -4.


PROOF

(x^2 + x +1)/(x+1) -x+4

((x+1)^2 - x) / (x+1) - x + 4

x+1 - x/(x+1) - x + 4

5 - x/(x+1)

finding limit as x reaches infinity of above term is 4