Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{\log 100+\log (0.01+x)}{x}$

Answer 1

HOPE IT HELPS U ✌️✌️

Answer 2

Answer:

$\rm \displaystyle \lim_{x \to 0}\ \frac{\log 100+\log (0.01+x)}{x} = 100$

Step-by-step explanation:

Since we have given $\rm \displaystyle \lim_{x \to 0}\ \frac{\log 100+\log (0.01+x)}{x}$

we know that log(100) = 2

so, apply limit x tends to 0 directly we get

$\dfrac{2 +log(0.01)}{0}$

we know that log(0.01) = -2,

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

We know that derivative of a constant number is zero.

$\dfrac{0+\dfrac{1}{0.01+x} }{1}$

So, now apply limit we get

$\dfrac{0+\dfrac{1}{0.01+0} }{1} = \dfrac{100}{1}$

So, $\rm \displaystyle \lim_{x \to 0}\ \frac{\log 100+\log (0.01+x)}{x} = 100$