Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{\log x+\log \Big(\frac{1+x}{x}\Big)}{x}$

Answer 1

HOPE IT HELPS U ✌️✌️✌️

Answer 2

Answer:

$\rm \displaystyle \lim_{x \to 0}\ \frac{\log x+\log \Big(\frac{1+x}{x}\Big)}{x} = \infty$

Step-by-step explanation:

Since we have given $\rm \displaystyle \lim_{x \to 0}\ \frac{\log x+\log \Big(\frac{1+x}{x}\Big)}{x}$

so, apply limit x tends to 0 directly we get

$\dfrac{log0 +log1}{0}$ which is $\dfrac{0}{0}$ form so, apply l'hospital rule i.e. differentiated each function with respect to its variable we get.

$\rm \displaystyle \lim_{x \to 0}\ \frac{\frac{1}{x} +\ \Big(\frac{1}{\frac{1+x}{x}}\Big)(\frac{-1}{x^{2} }) }{1}$

Now apply limit we get

$\rm \displaystyle \ \frac{\frac{1}{0} +\ \Big(\frac{1}{\frac{1+0}{0}}\Big)(\frac{-1}{0^{2} }) }{1}$

we get $\infty$

$\rm \displaystyle \lim_{x \to 0}\ \frac{\log x+\log \Big(\frac{1+x}{x}\Big)}{x} = \infty$