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Calculate the limit $\lim \limits_ {x \to \infty} \left(\frac{x^2+1}{x-1}\right)$

$\lim \limits_ {x \to \infty}\frac{x^2+1}{x-1}=\lim \limits_ {x \to \infty}\frac{x(x+\frac1x)}{x(1-\frac1x)}=\lim \limits_ {x \to \infty}\frac{x+\frac1x}{1-\frac1x}=\lim \limits_ {x \to \infty}\color{red}{\underbrace{\frac{\infty +0}{1-0}}_{\text{ not formal!}}}=\infty$

How do I express the marked part in a formal way? I know that adding something to infinity is wrong, because $\infty$ is not a number. Unfortunately, I don't have any ideas to correct this.

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  • $\begingroup$ What would you write for $\lim_{x\to\infty}(x+1)=?=\infty$ in the middle? $\endgroup$ – user587192 Nov 11 '18 at 0:00
  • $\begingroup$ Probably $\infty+1$, but that would be informal too and actually, this one is very obvious. $\endgroup$ – Doesbaddel Nov 11 '18 at 0:03
  • $\begingroup$ Then for "formal", I think you mean a "proof"? $\endgroup$ – user587192 Nov 11 '18 at 0:43
  • $\begingroup$ Yeah, that's what I'm trying to do. $\endgroup$ – Doesbaddel Nov 11 '18 at 19:13
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I often use

$$\ldots=\lim_ {x \to \infty}\frac{x+\frac1x}{1-\frac1x}=\left(\frac{\infty+0}{1-0}\right)=\infty$$

or directly as $x\to\infty$

$$\ldots=\frac{x+\frac1x}{1-\frac1x}\to \infty$$

In any case I suggest to avoid that one

$$\ldots=\lim_ {x \to \infty}\color{red}{\underbrace{\frac{\infty +0}{1-0}}_{\text{ not formal!}}}=\ldots$$

also in a not formal answer since we are writing the values assumed by the terms under the limit.

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From 2 onward 0 < 1 - 1/x < 1
Thus 2 < x < x/(1 - 1/x) < (1 + 1/x)/(1 - 1/x).
Desired conclusion follows.

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The idea is doing $$ \frac{x^2+1}{x-1}=\frac{x^2}{x}\frac{1+\dfrac{1}{x^2}}{1-\dfrac{1}{x}} $$ The limit of $$ \frac{1+\dfrac{1}{x^2}}{1-\dfrac{1}{x}} $$ is $1$ by standard rules on limits. Hence there is $M>0$ such that, for $x>M$, $$ \frac{1+\dfrac{1}{x^2}}{1-\dfrac{1}{x}}>\frac{1}{2} $$ Hence, for $x>M$, $$ \frac{x^2+1}{x-1}>\frac{x}{2} $$ Thus, given $K>0$, you can say that, for $x>\max\{M,2K\}$, you have $$ \frac{x^2+1}{x-1}>K $$ The same reasoning, with obvious changes, applies to every function of the form $$ \frac{a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0}{b_mx^n+b_{m-1}x^{m-1}+\dots+b_1x+b_0} $$ with $a_n\ne0$, $b_m\ne0$ and $n>m$.

If, instead, $n<m$, the limit is $0$. With $n=m$, the limit is $a_n/b_n$.

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