Question

plz guys help..... ​

Answer 1

As followed by hint, let,

$\quad$

$2-x=z$

$\quad$

Then,

$\quad$

$\bullet\quad x\to2\implies z\to0\\\\\bullet\quad 2-x=z\implies 3-x=1+z$

$\quad$

So,

$\quad$

$\displaystyle\mapsto\lim_{x\to 2}\dfrac {1}{(2-x)}\log (3-x)=\lim_{z\to0}\dfrac {1}{z}\log (1+z)\\\\\\\mapsto\lim_{x\to 2}\dfrac {1}{(2-x)}\log (3-x)=\lim_{z\to0}\dfrac {\log (1+z)}{z}$

$\quad$

It's a formula that,

$\quad$

$\large\boxed {\lim_{t\to0}\dfrac {\log (1+t)}{t}=1}$

$\quad$

Therefore,

$\quad$

$\displaystyle\mapsto\lim_{x\to 2}\dfrac {1}{(2-x)}\log (3-x)=\bf{\underline {\underline {1}}}$