$ \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?
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
indeed
$$\frac{x^2(1-1)-x(1-1)-(-1)+1}{x+1}=\frac{2}{x+1}\to 0$$
What you have seems fine (though maybe overkill); another way you can approach this problem is using long division to show that $$\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}.$$ The last term goes to $0$ as $x\to\infty$ while the first two terms combine with those in the original problem to get $(1-a)x-(1+b)\to0$ as $x\to\infty$ (but this is equivalent to what you already have).
The limit is: $$L=\lim_\limits{x\to \infty} ((2-2a)x-(a+b))$$ Consider the cases: $$L=\begin{cases} \infty, \ 2-2a\ne 0 \\ -(1+b), \ 2-2a=0\end{cases}$$ Further note: $$L=\begin{cases} 0, \ b=-1 \\ -(1+b)\ne 0, \ b \ne -1\end{cases}$$
Let $f(x)=\dfrac{x^2+1}{x+1}$. If $$ \lim_{x\to\infty}(f(x)-ax-b)=0 $$ then also $$ \lim_{x\to\infty}\frac{f(x)-ax-b}{x}=0 $$ Thus we must have $$ \lim_{x\to\infty}\left(\frac{x^2+1}{x(x+1)}-a\right)=0 $$ and therefore $a=1$. Now $$ \frac{x^2+1}{x+1}-x=\frac{x^2+1-x^2-x}{x+1}=\frac{-x}{x+1} $$ so $$ \lim_{x\to\infty}(f(x)-x-b)=-1-b $$ and therefore $b=-1$.
