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When I'm asked to find limits at infinity for a given rational function, what do I do? I'm aware of the result we obtain by comparing the degrees of the numerator and the denominator of a given rational function. I'm listing the questions below which I don't understand.

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

Here, a and b are some Real constants.

Answer to question 1: $a = 1$; $b = -1$. Answer to question 2: $a = 1$; $b = -3$. Answer to question 3: $a ≠ 1$; $b ∈ ℝ$.

In Q.1 since the value of limit is equal to 0, degree of the numerator must be less than that of the denominator. By using this I am able to get the value of a and b as 1 and -1 respectively. But for question 2 and 3 I'm unable to understand what happens.

If possible please tell me a good source to learn limits.

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  • $\begingroup$ If the degrees of the numerator and denominator are the same, then as $x \to \infty$, the ratio will tend to the ratio of the leading coefficients. You can prove this by dividing the numerator and denominator by the largest power of $x$. However I'm not so sure about the supposed answer to $3$. If $a = 2$, then it surely tends to $-\infty$. $\endgroup$ – Izaak van Dongen Jul 11 at 15:02
  • $\begingroup$ For question 2,if a=1 and b=3, it is ( 2x+3)/(x+1)= (2+3/x)/(1+1/x) as x tends to infinity , 1/x tends to 0, so lim (2+3/x)/(1+1/x) = 2 as x tends to infinity. $\endgroup$ – Subhajit Jul 11 at 15:06
  • $\begingroup$ For question 3, function becomes = ((1-a)x²-(a+b)x+(1-b))/(1+x) , since a not equal to 1, so, degree of numerator is strictly greater than the degree of denominator , so as x tends to infinity, limit of the given function goes to infinity. $\endgroup$ – Subhajit Jul 11 at 15:11
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    $\begingroup$ And also one thing here that if (1-a)<0 then the limit goes to -∞ $\endgroup$ – Subhajit Jul 11 at 15:18
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    $\begingroup$ @Izaak van Dongen Hmm you are right! You can see I add some! $\endgroup$ – Subhajit Jul 11 at 15:19
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I'll walk through how I would approach the first one (showing all work), and leave the rest up to you.

Find $a$ and $b$ such that $$\lim _{x\to\infty}\left(\frac{x^2+1}{x+1}-ax-b\right)=0.$$

I would first find a common denominator between all three terms and write it as one ratio: \begin{align} \lim_{x\to\infty}\left(\frac{x^{2} + 1 - ax(x+1) - b(x+1)}{x+1}\right) &= \lim_{x\to\infty}\left(\frac{x^{2} + 1 - ax^{2} - ax - bx - b}{x+1}\right)\\ &=\lim_{x\to\infty}\left(\frac{(1-a)x^{2}-(a+b)x+1-b}{x+1}\right) \end{align}

Dividing each term by the highest power of $x$ in the denominator:

\begin{align} \lim_{x\to\infty}\left(\frac{(1-a)x^{2}-(a+b)x+1-b}{x+1}\right) &= \lim_{x\to\infty}\left(\frac{(1-a)x - a - b + (1-b)x^{-1}}{1+x^{-1}}\right) \end{align}

Now, the $x^{-1}$ terms approach zero as $x\to \infty$, so we can rewrite this as

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

So, we're left with

$$-(a+b) +(1-a)\lim_{x\to\infty}x = 0$$

which implies that $a = 1$ so that $1-a = 0$ so we eliminate $\lim_{x\to\infty}x = \infty$. That then leaves us with $$-(1+b) = 0\implies b = -1.$$

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  • $\begingroup$ how did you take -(a+b) out? (in the second last step) $\endgroup$ – bhuvanesh Jul 11 at 16:21
  • $\begingroup$ @bhuvanesh the limit of a constant is just the constant. $\endgroup$ – DMcMor Jul 11 at 16:24
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  1. $$\dfrac{x^2+1}{x+1}=x-1+\frac2{x+1}\implies\dfrac{x^2+1}{x+1}-(x-1)=\frac2{x+1}\to0$$

  2. $$\dfrac{x^2-1}{x+1}=x-1\implies\dfrac{x^2-1}{x+1}-(x-3)=2\to2.$$

  3. $$\dfrac{x^2+1}{x+1}=x-1+\frac2{x+1}\implies\dfrac{x^2+1}{x+1}-ax-b=(1-a)x-1-b+\frac2{x+1}\\\to\pm\infty\ (a\ne-1).$$

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Let's rewrite the term to express it as a proper fraction by using the following method or long division.

\begin{align}\frac{x^2+1}{x+1}&=\frac{(x^2+x)-x+1}{x+1}, \text{add } x \text{ and subract } x \\&=\frac{x(x+1)-x-1+1+1}{x+1} , \text{add } -1 \text{ and subract } -1\\ &= \frac{(x-1)(x+1)+2}{x+1}\\ &= x-1 + \frac{2}{x+1}\end{align}

As $x \to \infty$, we have $\lim_{x \to \infty}\frac{2}{x+1}=0$, now let's study

$$\lim_{x \to \infty} x-1 + \frac{2}{x+1}-ax-b=\lim_{x \to \infty} [(1-a) x+(-1-b)]$$

There are three cases to consider:

Case $1$: $a < 1$, in that case $1-a$ is positive, and hence $(1-a)x+(-1-b)$, a line with positive slope will go to $\infty$. This is the third part that you want to solve.

Case $2$: $a>1$, in that case $1-a$ is negative, and hence $(1-a)x+(-1-b)$, a line with negative slope will go to $-\infty$.

Case $3$: $a=1$,

$$\lim_{x \to \infty} x-1 + \frac{2}{x+1}-ax-b=\lim_{x \to \infty} [(1-a) x+(-1-b)]=-1-b$$

Hence, if you want it to be equal to a real value $L$, you just let $-1-b=L$, that is $b=-1-L$ and $a=1$.

In the first part, $L=0$, we have $b=-1-0=-1$ and $a=1$.

In the second part, $L=2$, we have $b=-1-2=-3$ and $a=1$.


Overall takeaway, given an improper fraction and you are asked to study the limit, you might like to convert to proper fraction first and you can reduce the problem to taking limit of polynomial. In this case, just a straight line and the constant function.

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$$L=\lim _{x\to\infty}\left(\frac{x^2+1}{x+1}-ax-b\right)$$ $$\implies \lim _{x\to\infty}\left(x^2+1)(x(1+1/x)^{-1}-ax-b\right) $$ Use $(1+1/x)^{-1}=1-1/x+1/x^2+...$, when $x$ is very large then $$\implies \lim _{x\to\infty}[(x-1+2/x)-(ax+b)]= \lim_{x \to \infty}[(1-a)x+2/x-(b+1)]$$ Question (1): $L=0 \implies a=1,b=-1$

Question (2): $L=2 \implies a=1, b=-3$

Question (3): $L= +\infty \implies a<1$

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  • $\begingroup$ I wanna say one thing that is it really possible to say L= +∞ in the place of saying that L tends to +∞ without including extended real line! $\endgroup$ – Subhajit Jul 11 at 15:27
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To simplify the limit, make the change $t=x+1$: $$\lim _{x\to\infty}\left(\frac{x^2+1}{x+1}-ax-b\right)=\\ \lim _{t\to\infty}\left(\frac{(t-1)^2+1}{t}-a(t-1)-b\right)=\\ \lim _{t\to\infty}\left(t-2+\frac2t-at+a-b\right)=\\ \lim _{t\to\infty}\left((1-a)t+a-b-2\right)$$ Note: $\lim_\limits{t\to\infty} \frac2t=0$.

I) For the limit to be equal to $0$: $$\begin{align}&1) \ 1-a=0\Rightarrow a=1\\ &2) \ a-b-2=0 \Rightarrow 1-b-2=0\Rightarrow b=-1\end{align}$$ II) I assume the second limit also has $x^2+1$ on the numerator! (The given answer also hints to it). For the limit to be equal to $2$: $$\begin{align}&1) \ 1-a=0\Rightarrow a=1\\ &2) \ a-b-2=2\Rightarrow 1-b-2=2\Rightarrow b=-3\end{align}$$ III) For the limit to be equal to $\infty$ (that is, it does not exist): $$\begin{align}&1) \ 1-a\ne 0\Rightarrow a\ne 1\\ &2) \ b\in R\end{align}$$

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$Given$,

$$\lim _{x\to\infty}\left(\frac{x^2+1}{x+1}-ax-b\right)$$ and for the existince of the limit ,i.e it must me real,

$$\begin{align} \lim_{x\to\infty}\left(\frac{x^{2} + 1 - ax(x+1) - b(x+1)}{x+1}\right) &= \lim_{x\to\infty}\left(\frac{x^{2} + 1 - ax^{2} - ax - bx - b}{x+1}\right)\\ &=\lim_{x\to\infty}\left(\frac{(1-a)x^{2}-(a+b)x+1-b}{x+1}\right) \end{align}$$ For the existence of the limit, the degree of the numerator and the denominator must be the same .

$\begin{align} \lim_{x\to\infty}\left(\frac{(1-a)x^{2}-(a+b)x+1-b}{x+1}\right) &= \lim_{x\to\infty}\left(\frac{(1-a)x - a - b + (1-b)x^{-1}}{1+x^{-1}}\right) \end{align}$

= $\lim_{x \to \infty} [(1-a) x+(-1-b)]$

For the existence of the limit,

$case.1:$ $[(1-a) x-(a+b)]=0$

$a$ must be equal to $1$ for the existence of the limit

$\lim_{x \to \infty} [(1-1) x+(-1-b)]=0$

$Therefore$, $a=1$ , $b=-1$

Similarly,

$Case.2:$For, L=2, we have $b=−3$ and $a=1$.

$case.3:$For, $L=\infty$, we have $\begin{align}& a\ne 1\& \ b\in R\end{align}$

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Let me give you the good news here. You can solve such problems with your existing knowledge ("I am aware of the result...").

Let us then express this existing knowledge in concrete terms. Suppose we are given a rational function of $x$ and we are supposed to find its limit as $x\to\infty $. Then we have the following mutually exclusive and exhaustive possibilities :

  • If the degree of numerator is less than that of the denominator then the limit is $0$.
  • If the degree of numerator is equal to that of the denominator then the limit is non-zero and equal to the ratio of leading coefficients of the numerator and denominator.
  • If the degree of numerator is more than that of the denominator then the limit is $\infty$ or $-\infty$ depending on whether the ratio of leading coefficients of numerator and denominator is positive or negative.

Since the above possibilities are mutually exclusive and exhaustive each of the above statements is effectively of the "if and only if" variety instead of the weaker "if... then..." variety.

Now your three limit problems correspond to each of the three possibilities given above and you just need to rewrite the expression under limit as a rational function of $x$. Thus for the first problem the expression under limit can be written as $$\frac{(1-a)x^2-(a+b)x+1-b} {x+1} $$ Since the limit of above expression is given to be $0$ it follows that degree of numerator is less than that of denominator. Hence degree of numerator must be $0$ and coefficients of $x$ and $x^2$ in the numerator must vanish. This means that $$1-a=0,a+b=0$$ which gives us $$a=1,b=-1$$ For the second problem the expression under limit is $$\frac{(1-a)x^2-(a+b)x-1-b}{x+1}$$ and the given limit is $2$ so that the degree of both numerator and denominator must be same. It follows that degree of numerator is $1$ and hence the coefficient of $x^2$ in numerator must be $0$ ie $1-a=0$ or $a=1$. Further the limit $2$ also equals the ratio of leading coefficients of numerator and denominator so that $$2=-\frac{a+b}{1}$$ and then $b=-3$.

Proceed in same fashion and solve the third and last problem. The correct solution is $a<1,b\in\mathbb {R} $ and I suppose your question has a typo in giving the solution to third problem.


You might also be interested to know what happens when $x\to-\infty $. Well we don't need to make three more rules for this. Just put $x=-t$ and transform the problem to $t\to\infty $ and apply the same rules.

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