Question

Evaluate the limit:
$\underset{x\to 2}{lim}\frac{x-2}{{\log }_a(x-1)}$

Answer 1

Given:
$\underset{x\to 2}{lim}\frac{x-2}{{\log }_a(x-1)}$

Put $x=2+h$
If $x\to 2$, then $h\to 0$

$=\underset{h\to 0}{lim}\frac{2+h-2}{{\log }_a(2+h-1)}$

$=\underset{h\to 0}{lim}\frac{h}{{\log }_a(1+h)}$

$=\underset{h\to 0}{lim}\frac{h}{\frac{{\log }_e(1+h)}{{\log }_{e}a}}[\because {\log }_{a}b=\frac{{\log }_{e}b}{{\log }_{e}a}]$

$=\underset{h\to 0}{lim}\frac{h{\log }_{e}a}{{\log }_e(1+h)}$

$=\underset{h\to 0}{lim}\frac{{\log }_{e}a}{\frac{{\log }_e(1+h)}{h}}$

$=\frac{{\log }_{e}a}{1}=\log a$
$[\because \underset{x\to 0}{lim}\frac{{\log }_e(1+x)}{x}=1]$

Therefore,
$\underset{x\to 2}{lim}\frac{x-2}{{\log }_a(x-1)}=\log a$