Question

Find the value of $\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\log \left( {1 + 9x} \right)}}{x}} \right]$

Answer 1

Hint: We can see this is $\dfrac{0}{0}$ form then we can use L’ Hospital rule in which we differentiate numerator and denominator with respect to x and check whether any form of limit is coming or not if not we can substitute x=0. And get the value of the limit.

Complete step-by-step answer:
 $\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\log \left( {1 + 9x} \right)}}{x}} \right]$
Using l’ hospitals rule
$\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\dfrac{{d\log \left( {1 + 9x} \right)}}{{dx}}}}{{\dfrac{{dx}}{{dx}}}}} \right]$
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\dfrac{9}{{1 + 9x}}}}{1}} \right]$
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{9}{{1 + 9x}}} \right]$
When we substitute x=0 then limit has a finite value
$\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\log \left( {1 + 9x} \right)}}{x}} \right] = 9$

Note: If $\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)$then we can find the limit of the given question left hand limit.
Alternative method :-
It is known as $\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\log \left( {1 + x} \right)}}{x}} \right] = 1$
Multiply numerator and denominator by 9 we get
$\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{9\log \left( {1 + 9x} \right)}}{{9x}}} \right]$=9 $\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\log \left( {1 + 9x} \right)}}{{9x}}} \right]$=9