For one of the example problems from lecture, we were asked to show that for any $a \neq 1$ from the definition of a limit that the limit as $x$ approaches $a$ for $f(x)=x^2/(x-1)=a^2/(a-1)$.
after the $|f(x)-L|$ and merging the fractions together, she used the triangle inequality to obtain $((|a||x|+|a|+|x|)/(|x-1||a-1|)) |x-a|$ from $|(ax^2-x^2-a^2x+a^2)/(x-1)(a-1)|$
I am a bit confused on what was defined as $|a|$ and $|b|$ for the triangle inequality (|a+b| <= |a| + |b|). In other words, can anyone help explain how the equation was broken up? thanks
Factoring by parts we have
$\left|\frac{(ax^2-x^2-a^2x+a^2)}{(x-1)(a-1)}\right| = \left|\frac{(x-a)(ax-a-x)}{(x-1)(a-1)}\right| = |x-a| \frac{|ax-a-x|}{|x-1||a-1|} \le |x-a| \frac{|a||x|+|a|+|x|}{|x-1||a-1|} $.
The triangle inequality is used to obtain
$|ax-a-x| \le |a||x|+|a|+|x|$