Alg7.1.5

first we need to point out that if $x = 1$ we have division

by 0 wich we dont want and also it can be easily seen that

if $x \in (1,3)$ the inequality will also not be true.

hence,we consider the above as part of the statement of the problem and

continue to solve it.

We have to show that :

$\left|\frac{x+1}{x-1}\right| \leq 2$

but,

$\left|\frac{x+1}{x-1}\right| = \left|\frac{x-1 +2}{x-1}\right| = \left|1 +\frac{2}{x-1}\right|$

so now we use the triangel inequality $|a+b|\leq |a| + |b|$

and we get that $\left|1 +\frac{2}{x-1}\right|\leq 1 + \left|\frac{2}{x-1}\right|$

but as we said before that $x > 3, x < 2$

we can conclude that $\frac{2}{x-1} \leq 1$

we are done with this case.

now let us consider $x < 1, x > - 1$ we will find that

the denominator will be negative,the numerator will be positive,but much less

than the denominator thus the fraction will be less than one and by modulus it is

for sure less than 2.

for x less than 1 it is obvious that the fraction is equivalent to one without

the modulus

so our problem is done