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

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.

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

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.

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

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