Find the value of \[\mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\left( {2x - 3} \right)\left( {3x{\text{ }} - {\text{ }}4} \right)}}{{\left( {4x - 5} \right)\left( {5x - 6} \right)}}\]
A. \[\dfrac{1}{10}\]
B. \[0\]
C. \[\dfrac{1}{5}\]
D. \[\dfrac{3}{10}\]

Hint: In this question, we need to find the value of the limit\[\mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\left( {2x - 3} \right)\left( {3x{\text{ }} - {\text{ }}4} \right)}}{{\left( {4x - 5} \right)\left( {5x - 6} \right)}}\]. For this, we will divide numerator and denominator by $x^2$ and simplify the limit to get the desired result.

Complete step-by-step solution:
We know that \[\mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\left( {2x - 3} \right)\left( {3x{\text{ }} - {\text{ }}4} \right)}}{{\left( {4x - 5} \right)\left( {5x - 6} \right)}}\]
Let \[L = \mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\left( {2x - 3} \right)\left( {3x{\text{ }} - {\text{ }}4} \right)}}{{\left( {4x - 5} \right)\left( {5x - 6} \right)}}\]
Let us simplify this.
By dividing \[{x^2}\] to the numerator and the denominator, we get
\[L = \mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\dfrac{{\left( {2x - 3} \right)\left( {3x{\text{ }} - {\text{ }}4} \right)}}{{{x^2}}}}}{{\dfrac{{\left( {4x - 5} \right)\left( {5x - 6} \right)}}{{{x^2}}}}}\]
By separating the denominator, we get
\[L = \mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\dfrac{{\left( {2x - 3} \right)}}{x}\dfrac{{\left( {3x{\text{ }} - {\text{ }}4} \right)}}{x}}}{{\dfrac{{\left( {4x - 5} \right)}}{x}\dfrac{{\left( {5x - 6} \right)}}{x}}}\]
By simplifying, we get
\[L = \mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\left( {\dfrac{{2x}}{x} - \dfrac{3}{x}} \right)\left( {\dfrac{{3x}}{x} - \dfrac{4}{x}} \right)}}{{\left( {\dfrac{{4x}}{x} - \dfrac{5}{x}} \right)\left( {\dfrac{{5x}}{x} - \dfrac{6}{x}} \right)}}\]
\[L = \mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\left( {2 - \dfrac{3}{x}} \right)\left( {3 - \dfrac{4}{x}} \right)}}{{\left( {4 - \dfrac{5}{x}} \right)\left( {5 - \dfrac{6}{x}} \right)}}\]
By putting\[x = \infty \], in the above equation, we get
\[L = \;\dfrac{{\left( {2 - \dfrac{3}{\infty }} \right)\left( {3 - \dfrac{4}{\infty }} \right)}}{{\left( {4 - \dfrac{5}{\infty }} \right)\left( {5 - \dfrac{6}{\infty }} \right)}}\]
Let us simplify it further.
So, we get
\[L = \;\dfrac{{\left( {2 - 3\left( {\dfrac{1}{\infty }} \right)} \right)\left( {3 - 4\left( {\dfrac{1}{\infty }} \right)} \right)}}{{\left( {4 - 5\left( {\dfrac{1}{\infty }} \right)} \right)\left( {5 - 6\left( {\dfrac{1}{\infty }} \right)} \right)}}\]
But we know that \[\dfrac{1}{\infty } = 0\]
Thus, we get
\[L = \;\dfrac{{\left( {2 - 3\left( 0 \right)} \right)\left( {3 - 4\left( 0 \right)} \right)}}{{\left( {4 - 5\left( 0 \right)} \right)\left( {5 - 6\left( 0 \right)} \right)}}\]
\[L = \;\dfrac{{\left( {2 - 0} \right)\left( {3 - 0} \right)}}{{\left( {4 - 0} \right)\left( {5 - 0} \right)}}\]
\[L = \;\dfrac{{\left( 2 \right)\left( 3 \right)}}{{\left( 4 \right)\left( 5 \right)}}\]
By simplifying it further, we get
\[L = \;\dfrac{6}{{20}}\]
\[L = \;\dfrac{3}{{10}}\]
That is, we can say that
\[\mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\left( {2x - 3} \right)\left( {3x{\text{ }} - {\text{ }}4} \right)}}{{\left( {4x - 5} \right)\left( {5x - 6} \right)}} = \dfrac{3}{{10}}\]
Hence, the value of the limit \[\mathop {\lim }\limits_{x \to \infty } \;\dfrac{{\left( {2x - 3} \right)\left( {3x{\text{ }} - {\text{ }}4} \right)}}{{\left( {4x - 5} \right)\left( {5x - 6} \right)}}\] is \[\dfrac{3}{{10}}\].
Therefore, the correct option is (D).

Note: Many students generally make mistakes in solving the limit. They make take infinity for the ratio of \[\dfrac{1}{\infty }\]. If so, they may get the wrong end result. Here, the main trick to solve this question is to divide the numerator and the denominator of the given limit by \[{x^2}\] to get the correct result.