<s>[INST] $ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $ $ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $
Answer:
$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \\ \Rightarrow  \lim_{x \to \infty} [\frac{x^2+1-ax^2-ax-bx-b}{x+1}]=0 \\ \Rightarrow \lim_{x \to \infty} \frac{2x-2ax-a-b}{1}=0 \\ \Rightarrow 2x-2ax-a-b=0  \ \ (?)  $ 
Comparing both sides , we get 
$ 2-2a=0 \\ a+b=0 \ $
Solving , we get 
$ a=1 , \ b=-1 \ $
But I am not sure about the above line where question mark is there.
Can you help me?
 [/INST] Note that
$$\frac{x^2+1}{x+1}-ax-b=\frac{x^2+1-ax^2-bx-ax-b}{x+1}=\frac{x^2(1-a)-x(a+b)-b+1}{x+1}$$
and in order to have limi zero we need


*

*$(1-a)=0 \implies a=1$

*$(a+b)=0\implies b=-1$


indeed
$$\frac{x^2(1-1)-x(1-1)-(-1)+1}{x+1}=\frac{2}{x+1}\to 0$$
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