Solve the following: $\mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + 4h} \right) - 2{{\log }_e}\left( {1 + 2h} \right)}}{{{h^2}}}$
A) $ - 4$
B) 4
C) 2

Hint:
Here, we will apply L’Hopital’s rule to reduce the limit. We will then simplify the expression by taking the LCMs and cancelling out the like terms. When the limit has been simplified to an extent such that if we substitute $h = 0$ then our denominator does not turn out to be 0, then, we will substitute $h = 0$ and remove the limit to find the required value of $h$ which satisfies this limit.

Formula Used:
We will use the following formulas:
1) $\dfrac{{dy}}{{dx}}{x^n} = n{x^{n - 1}}$
2) $\dfrac{{dy}}{{dx}}\log x = \dfrac{1}{x}$

Complete step by step solution:
The given expression is $\mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + 4h} \right) - 2{{\log }_e}\left( {1 + 2h} \right)}}{{{h^2}}}$.
Now, in order to solve the given limit further, we will apply the L’Hopital’s rule. According to this rule if we have an indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, we differentiate the numerator and the denominator and then we take the limit.
Now, applying L’Hopital’s rule and differentiating the expression using $\dfrac{{dy}}{{dx}}{x^n} = n{x^{n - 1}}$ and $\dfrac{{dy}}{{dx}}\log x = \dfrac{1}{x}$, we get
$\mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + 4h} \right) - 2{{\log }_e}\left( {1 + 2h} \right)}}{{{h^2}}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{4}{{1 + 4h}} - 2\left( {\dfrac{2}{{1 + 2h}}} \right)}}{{2h}}$
Multiplying the terms, we get
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + 4h} \right) - 2{{\log }_e}\left( {1 + 2h} \right)}}{{{h^2}}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{4}{{1 + 4h}} - \left( {\dfrac{4}{{1 + 2h}}} \right)}}{{2h}}$
Solving further by taking LCM in the denominator of the fractions present in the numerator, we get
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + 4h} \right) - 2{{\log }_e}\left( {1 + 2h} \right)}}{{{h^2}}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{4 \times \dfrac{{1 + 2h - 1 - 4h}}{{\left( {1 + 4h} \right)\left( {1 + 2h} \right)}}}}{{2h}}\]
Simplifying the expression, we get
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + 4h} \right) - 2{{\log }_e}\left( {1 + 2h} \right)}}{{{h^2}}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{4 \times \left( { - 2h} \right)}}{{2h\left( {1 + 4h} \right)\left( {1 + 2h} \right)}}\]
Cancelling out the same terms from the numerator as well as the denominator, we get,
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + 4h} \right) - 2{{\log }_e}\left( {1 + 2h} \right)}}{{{h^2}}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 4}}{{\left( {1 + 4h} \right)\left( {1 + 2h} \right)}}\]
Now, substituting $h = 0$ and hence, removing the limit, we get,
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + 4h} \right) - 2{{\log }_e}\left( {1 + 2h} \right)}}{{{h^2}}} = \dfrac{{ - 4}}{{\left( {1 + 4 \times 0} \right)\left( {1 + 2 \times 0} \right)}} = - 4\]

Therefore, the required value of $\mathop {\lim }\limits_{h \to 0} \dfrac{{{{\log }_e}\left( {1 + 4h} \right) - 2{{\log }_e}\left( {1 + 2h} \right)}}{{{h^2}}} = - 4$
Hence, option A is the correct answer.

Note:
In mathematics, a limit is a value that a function approaches as the input or the index approaches some value. Limits are an essential element of calculus and are used to define continuity, integrals and derivatives. The idea of limits was first developed by ‘Archimedes of Syracuse’ to measure curved figures and the volume of a sphere in the third century B.C. Hence, limits are being used for such a long time.