Question

# The value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{\log \left( {1 + 2x} \right)}}{x}$ is equal to:${\text{A}}{\text{. 1}} \\ {\text{B}}{\text{. 2}} \\ {\text{C}}{\text{. 3}} \\ {\text{D}}{\text{. }}\dfrac{3}{2} \\ {\text{E}}{\text{. }}\dfrac{1}{2} \\$

Hint: Here, to solve this question, the first method should be to apply the limit. Since, we get an undetermined form we have to use L’Hospital’s rule to solve the question and chain rule has to be applied for the differentiation,

Here in this equation take the limit of numerator and denominator and evaluate it$= \dfrac{{\mathop {\lim }\limits_{x \to 0} {\text{ }}\log \left( {1 + 2x} \right)}}{{\mathop {\lim {\text{ }}x}\limits_{x \to 0} }} \\ = \dfrac{{\log \left( {1 + 2\mathop {\lim }\limits_{x \to 0} x} \right)}}{{\mathop {\lim }\limits_{x \to 0} x}} \\ = \dfrac{{\log \left( {1 + 2 \times 0} \right)}}{{\mathop {\lim }\limits_{x \to 0} x}} \\ = \dfrac{{\log \left( 1 \right)}}{{\mathop {\lim }\limits_{x \to 0} x}} \\$
The natural logarithm of $1$ is $0$. Therefore, we get
$= \dfrac{0}{0}$
Since it is undefined, we have to apply L’Hospital’s Rule which states that, if $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ is in an indeterminate form e.g. $\dfrac{0}{0}or\dfrac{\infty }{\infty }$ ,then we have to solve this using $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ and then solve the limit which implies that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
By using L’Hospital’s rule we will get,

$\dfrac{{\mathop {\lim }\limits_{x \to 0} {\text{ }}\log \left( {1 + 2x} \right)}}{{\mathop {\lim {\text{ }}x}\limits_{x \to 0} }} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left[ {\log \left( {1 + 2x} \right)} \right]}}{{\dfrac{d}{{dx}}\left[ x \right]}}$, where $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to 0} {\text{ }}\log \left( {1 + 2x} \right)}}{{\mathop {\lim {\text{ }}x}\limits_{x \to 0} }}$ and
$\mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left[ {\log \left( {1 + 2x} \right)} \right]}}{{\dfrac{d}{{dx}}\left[ x \right]}}$
So, now we have to operate only R.H.S
Now, we will differentiate it by using chain rule in numerator i.e. $\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$, where $\dfrac{d}{{dx}}\left[ {\log \left( {1 + 2x} \right)} \right]$ can be written as $\dfrac{d}{{dx}}\left[ {\log \left( {1 + 2x} \right)} \right] \times \dfrac{d}{{dx}}\left( {1 + 2x} \right)$
.As we know, derivative of $\log x = \dfrac{1}{x}$ and then we will derivate the function in $\log$ as well to get
$\Rightarrow \dfrac{1}{{1 + 2x}} \times 2$
Since, we get the numerator and we know the value of $\dfrac{d}{{dx}}\left[ x \right] = 1$ i.e. the value of denominator
So the original equation becomes,
$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{1 + 2x}}}}{1} \times 2$
$= \mathop {\lim }\limits_{x \to 0} \dfrac{2}{{1 + 2x}}$
Solving the limit
$= \dfrac{2}{{1 + 2 \times 0}} \\ = 2 \\$
So, the correct answer is ${\text{B}}{\text{. 2}}$

Note: L’Hospital’s rule is the most prominent part of the question as it guides us to proceed further towards our next crucial step i.e. Chain rule to $\log$, which needs to be applied carefully as there is a chance of error and do remember to differentiate the $\log$ and function given inside the $\log$. By using these rules, we will be able to solve this question and find the correct answer.