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Hi I have a question regarding of limits to infinity please help which I need to find the constant number for a and b. Please help! Thank You!

The question states the user to find the following constants a and b:

$$\lim_{x\to\infty}\left(\frac {x^2 + 1}{x+1}-ax-b\right)=0$$

Thank You!!!

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the limit not should be in x? –  rlartiga Mar 11 at 18:49
    
I am also not sure though... –  AskingQnsPro Mar 11 at 19:12

5 Answers 5

up vote 1 down vote accepted

Combining yields $$\lim_{x\to\infty}\frac{x^2-ax^2-(a+b)x+1-b}{x+1}\to 0$$

which holds when the numerator is any constant. What $a,b$ values can you choose so the $x$ terms in the numerator cancel out?

We want to find $a,b\in\mathbb{R}$ such that $(1−a)x^2−(a+b)x+(1−b)=0x^2+0x+c$.

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I tried using simultaneous equations yet a and b still cannot be found. Is there a way to equate the numerator into = 0? –  AskingQnsPro Mar 12 at 13:29
    
If the numerator is a constant, then the $x^2$ coefficient is zero. Same with the x coefficient. So the numerator is $(1−a)x^2−(a+b)x+(1−b)=0x^2+0x+c$ –  Joshua Biderman Mar 12 at 18:22

Hint: Note that by polynomial division, or otherwise, $$\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}.$$

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Hint: Write everything as one fraction (with common denominator). The term $x^2$ should vanish from the numerator, otherwise the whole expression would diverge to infinity. Moreover the term $x$ should also vanish from the numerator, otherwise the whole expression would converge to a certain constant. This will give two equations for the two unknowns $a$ and $b$.

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$$\lim_{x\to\infty}\left(\frac {x^2 + 1}{x+1}-ax-b\right)=0$$ $$\lim_{x\to\infty}\left(\frac {x^2 + 1-ax^2-ax-bx-b}{x+1}\right)=0$$ $$\lim_{x\to\infty}\left(\frac {(1-a)x^2 + -(a+b)x+(1-b)}{x+1}\right)=0$$

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Hint: Rewrite $\frac{x^2 + 1}{x+1}-ax-b$ as $$ \frac{x^2 + 1-ax^2-ax-bx-b}{x+1} = \frac{(1-a)x^2 -(a+b)x + 1-b}{x+1} $$ and argue that in order for the limit when $x\to\infty$ to be $0$, both the coefficients of $x^2$ and $x$ in the numerator must be $0$ (why?).

This will give you $a$ and $b$.

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