# function limit with parameters.

Find $a,b$ such that:

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

I have no idea how to solve this exercise. I know how to prove that the limit is $0$ without parameters $a , b.$

• No it isn't, it's $+\infty.$. It only becomes 0 for a specific (nonzero) choice of $a.$ – Justpassingby Dec 16 '15 at 12:56

## 5 Answers

It's not hard to see that the expression is :

$$\frac{x^2(1-a)+x(-a-b)+(1-b)}{x+1}$$

Now you know that the limit of the quotient of two polynomials is $0$ only when the denominator has greater degree than the numerator .

This means that the numerator has degree $0$ so the $x^2$ and $x$ coefficients must vanish .

$$1-a=0$$ and also : $$-a-b=0$$

Thus the numbers are $a=1$ and $b=-1$ .

• Because if the polynomial has degree $0$ than all terms of higher degree must vanish . This means that their respective coefficients vanish . – user252450 Dec 16 '15 at 13:43
• thanks you ! but why you decide that the numerator has to be 0 ? it's only one option I think... maybe it can be smaller than the denominator and != 0 ? – Silas2033 Dec 16 '15 at 13:44
• I'm not saying the numerator must be $0$ . The numerator is of degree $2$ and the denominator of degree $1$ but for the limit to be $0$ the numerator must have a lower degree . This means that the numerator has degree $0$ so the coefficients of the terms $x$ and $x^2$ must vanish . – user252450 Dec 16 '15 at 13:46
• thank you. I wish you were my teacher at college – Silas2033 Dec 16 '15 at 13:50

From the beginning: Let $$f(x)=\frac{x^2+1}{x+1}- a x-b,$$

$$\lim_{x\to+\infty} f(x) = (1-a) \infty,$$ so we have $a \neq 1 \implies \nexists \lim_{x\to+\infty} f(x).$ So we know $a=1$, which yields: $$\lim_{x\to+\infty} f(x) = -1-b \qquad \text{for } a=1.$$ Therefore, $(a,b)=(1,-1)$ is the only tuple in $\mathbb{R}^2$ that satisfies $$\lim_{x\to+\infty} \frac{x^2+1}{x+1}- a x-b = 0.$$

Put the entire expression on the same denominator, and choose $a$ and $b$ such that the polynomial in the numerator has the lowest possible degree.

• thank u for your post. what do you mean lowest possible degree? – Silas2033 Dec 16 '15 at 13:03
• First choose $a$ such that the term of the highest degree cancels. Then try to choose $b$ such that the term of the highest-but-one degree cancels as well (if possible). If that works then what you are left with (in the denominator) is a constant. – Justpassingby Dec 16 '15 at 13:06

Hint: $$ax+b=(ax+b)\frac{x+1}{x+1}=\frac{ax^2+(a+b)x+b}{x+1},$$ such that $$\frac{x^2+1}{x+1}-(ax+b)=\frac{x^2+1-\left(ax^2+(a+b)x+b\right)}{x+1}.$$ Can you continu from here?

• sorry I didn't understand the hint – Silas2033 Dec 16 '15 at 13:05
• The enumerator simplifies to $(1-a)x^2+(-a-b)x+1-b$. Write this as $\alpha x^2+\beta x+\gamma$. So you want $\lim_{x\to\infty}{\frac{\alpha x^2+\beta x+\gamma}{x+1}}=0$. Now, what does that say about $\alpha$, $\beta$ and $\gamma$? – Eric S. Dec 16 '15 at 13:10
• Also, if you want to figure it out yourself, don't look at the other answers XD – Eric S. Dec 16 '15 at 13:11

note that $$\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}$$