proving the limit of a divergent function

Question

$\lim \limits_{x \to 1_+}\frac{x+1}{x-1}=\infty$

Pf: $\forall M>0$, $\exists \delta=\delta(M,x_0)>0$ such that $f(x)>M$ for all $x$ with $0<|x-x_0|<\delta$

So $\frac{x+1}{x-1}>M$

$M(x-1)<x+1$

$Mx-M<x+1$

$Mx-x<M+1$

$x(M-1)<M+1$

$x<\frac{M+1}{M-1}$

Therefore if $1<x<1+\delta$

Fix $M>0$, if $x>1$, then $\frac{x+1}{x-1}>\frac{x+1}{1}>x+1>M$ so set $\delta=1+M$

Answer 1

You have done it almost right. At the end, it suffices to say you should choose a $\delta $ such that

$$1 < x < 1 + \delta \le {{M + 1} \over {M - 1}}$$

and hence any $\delta $ satisfying

$$\delta \le {{M + 1} \over {M - 1}} - 1 = {2 \over {M - 1}}$$

will do the trick. The simplest one is to take $\delta = {2 \over {M - 1}}$.