Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{\log(3-x)-\log(3+x)}{x}$

Answer 1

HOPE IT HELPS U ✌️✌️

Answer 2

Answer:

$\rm \displaystyle \lim_{x \to 0}\ \frac{\log(3-x)-\log(3+x)}{x} = \frac{-2}{3}$

Step-by-step explanation:

Since we have given $\rm \displaystyle \lim_{x \to 0}\ \frac{\log(3-x)-\log(3+x)}{x}$

So, apply limit x tends to 0 directly we get,$\dfrac{log3-log3}{0}$

So,we get  $\dfrac{0}{0}$ form so, apply l'hospital rule i.e. differentiated each function with respect to its variable we get.

$\dfrac{\frac{-1}{3-x}-\frac{1}{3+x}}{1}$

Now apply limit , we get,

$\dfrac{\dfrac{-1}{3}-\dfrac{1}{3}}{1} = \dfrac{-2}{3}$

$\rm \displaystyle \lim_{x \to 0}\ \frac{\log(3-x)-\log(3+x)}{x} = \frac{-2}{3}$