Question

Find the value of the limit \[\mathop {lim}\limits_{x \to \infty } \left[ {\left( {\dfrac{{{x^3}}}{{\left( {3{x^2} - 4} \right)}}} \right) - \left( {\dfrac{{{x^2}}}{{\left( {3x + 2} \right)}}} \right)} \right]\]
A. \[ - \left( {\dfrac{1}{4}} \right)\]
B. \[ - \left( {\dfrac{1}{2}} \right)\]
C. 0
D. \[\dfrac{2}{9}\]

Answer 1

Hint: In the given question, the equation of limit is given. We need to simplify the terms present in the brackets. We will divide the denominator and numerator by \[{x^3}\]. Then we will put \[x = \infty \] and substitute \[\dfrac{1}{\infty } = 0\] to get the value of given limit.
Formula Used:
\[\dfrac{a}{b} - \dfrac{c}{d} = \dfrac{{ad - bc}}{{bd}}\
\[\dfrac{n}{\infty } = 0\] , where \[n\] is any number.
Complete step by step solution:
The given equation is, \[\mathop {lim}\limits_{x \to \infty } \left[ {\left( {\dfrac{{{x^3}}}{{\left( {3{x^2} - 4} \right)}}} \right) - \left( {\dfrac{{{x^2}}}{{\left( {3x + 2} \right)}}} \right)} \right]\].
Let \[L\] be the value of the limit.
\[L = \mathop {lim}\limits_{x \to \infty } \left[ {\left( {\dfrac{{{x^3}}}{{\left( {3{x^2} - 4} \right)}}} \right) - \left( {\dfrac{{{x^2}}}{{\left( {3x + 2} \right)}}} \right)} \right]\]
Simplify the above equation.
\[L = \mathop {lim}\limits_{x \to \infty } \left[ {\dfrac{{{x^3}\left( {3x + 2} \right) - {x^2}\left( {3{x^2} - 4} \right)}}{{\left( {3{x^2} - 4} \right)\left( {3x + 2} \right)}}} \right]\]
\[ \Rightarrow \]\[L = \mathop {lim}\limits_{x \to \infty } \left[ {\dfrac{{3{x^4} + 2{x^3} - 3{x^4} + 4{x^2}}}{{9{x^3} + 6{x^2} - 12x - 8}}} \right]\]
\[ \Rightarrow \]\[L = \mathop {lim}\limits_{x \to \infty } \left[ {\dfrac{{2{x^3} + 4{x^2}}}{{9{x^3} + 6{x^2} - 12x - 8}}} \right]\]
\[ \Rightarrow \]\[L = \mathop {lim}\limits_{x \to \infty } \left[ {\dfrac{{2{x^2}\left( {x + 2} \right)}}{{9{x^3} + 6{x^2} - 12x - 8}}} \right]\]
\[ \Rightarrow \]\[L = 2\mathop {lim}\limits_{x \to \infty } \left[ {\dfrac{{{x^2}\left( {x + 2} \right)}}{{9{x^3} + 6{x^2} - 12x - 8}}} \right]\]
Divide the denominator and numerator by \[{x^3}\].
\[ \Rightarrow \]\[L = 2\mathop {lim}\limits_{x \to \infty } \left[ {\dfrac{{1 + \dfrac{2}{x}}}{{9 + \dfrac{6}{x} - \dfrac{{12}}{{{x^2}}} - \dfrac{8}{{{x^3}}}}}} \right]\]
\[ \Rightarrow \]\[L = 2\left[ {\dfrac{{1 + \dfrac{2}{\infty }}}{{9 + \dfrac{6}{\infty } - \dfrac{{12}}{{{\infty ^2}}} - \dfrac{8}{{{\infty ^3}}}}}} \right]\]
Apply the formula \[\dfrac{n}{\infty } = 0\] , where \[n\] is any number.
\[ \Rightarrow \]\[L = 2\left[ {\dfrac{{1 + 0}}{{9 + 0 - 0 - 0}}} \right]\]
\[ \Rightarrow \]\[L = \dfrac{2}{9}\]

Hence the correct option is option D
Note: Students are often confused with the value \[\dfrac{n}{\infty } = 0\] and \[\dfrac{n}{\infty } = \infty\,\ or\,\ undefined\] where \[n\] is any number. But the correct value is \[\dfrac{n}{\infty } = 0\].