JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 12)
$$\mathop {\lim }\limits_{x \to a} {{{{\left( {a + 2x} \right)}^{{1 \over 3}}} - {{\left( {3x} \right)}^{{1 \over 3}}}} \over {{{\left( {3a + x} \right)}^{{1 \over 3}}} - {{\left( {4x} \right)}^{{1 \over 3}}}}}$$ ($$a$$ $$ \ne $$ 0) is equal to :
$$\left( {{2 \over 9}} \right){\left( {{2 \over 3}} \right)^{{1 \over 3}}}$$
$$\left( {{2 \over 3}} \right){\left( {{2 \over 9}} \right)^{{1 \over 3}}}$$
$${\left( {{2 \over 3}} \right)^{{4 \over 3}}}$$
$${\left( {{2 \over 9}} \right)^{{4 \over 3}}}$$
Explanation
L = $$\mathop {\lim }\limits_{x \to a} {{{{\left( {a + 2x} \right)}^{{1 \over 3}}} - {{\left( {3x} \right)}^{{1 \over 3}}}} \over {{{\left( {3a + x} \right)}^{{1 \over 3}}} - {{\left( {4x} \right)}^{{1 \over 3}}}}}$$
$$= \mathop {\lim }\limits_{h \to 0} {{{{(a + 2(a + h))}^{1/3}} - {{(3(a + h))}^{1/3}}} \over {{{(3a + a + h)}^{1/3}} - {{(4(a + h))}^{1/3}}}}$$
= $$\mathop {\lim }\limits_{h \to 0} {{{{(3a)}^{1/3}}{{\left( {1 + {{2h} \over {3a}}} \right)}^{1/3}} - {{(3a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}} \over {{{(4a)}^{1/3}}{{\left( {1 + {h \over {4a}}} \right)}^{1/3}} - {{(4a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}}}$$
= $$\mathop {\lim }\limits_{h \to 0} \left( {{{{3^{1/3}}} \over {{4^{1/3}}}}} \right)\left[ {{{\left( {1 + {{2h} \over {9a}}} \right) - \left( {1 + {h \over {3a}}} \right)} \over {\left( {1 + {h \over {12a}}} \right) - \left( {1 + {h \over {3a}}} \right)}}} \right]$$
$$ = {\left( {{3 \over 4}} \right)^{1/3}}{{\left( {{2 \over 9} - {1 \over 3}} \right)} \over {\left( {{1 \over {12}} - {1 \over 3}} \right)}} = {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{8 - 12} \over {3 - 12}}} \right)$$
$$ = {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{ - 4} \over { - 9}}} \right) = {{{4^{1 - {1 \over 3}}}} \over {{3^{2 - {1 \over 3}}}}} = {{{4^{2/3}}} \over {{3^{5/3}}}}$$
$$ = {{{{(8 \times 2)}^{1/3}}} \over {{{(27 \times 9)}^{1/3}}}} = {2 \over 3}{\left( {{2 \over 9}} \right)^{1/3}}$$
$$= \mathop {\lim }\limits_{h \to 0} {{{{(a + 2(a + h))}^{1/3}} - {{(3(a + h))}^{1/3}}} \over {{{(3a + a + h)}^{1/3}} - {{(4(a + h))}^{1/3}}}}$$
= $$\mathop {\lim }\limits_{h \to 0} {{{{(3a)}^{1/3}}{{\left( {1 + {{2h} \over {3a}}} \right)}^{1/3}} - {{(3a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}} \over {{{(4a)}^{1/3}}{{\left( {1 + {h \over {4a}}} \right)}^{1/3}} - {{(4a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}}}$$
= $$\mathop {\lim }\limits_{h \to 0} \left( {{{{3^{1/3}}} \over {{4^{1/3}}}}} \right)\left[ {{{\left( {1 + {{2h} \over {9a}}} \right) - \left( {1 + {h \over {3a}}} \right)} \over {\left( {1 + {h \over {12a}}} \right) - \left( {1 + {h \over {3a}}} \right)}}} \right]$$
$$ = {\left( {{3 \over 4}} \right)^{1/3}}{{\left( {{2 \over 9} - {1 \over 3}} \right)} \over {\left( {{1 \over {12}} - {1 \over 3}} \right)}} = {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{8 - 12} \over {3 - 12}}} \right)$$
$$ = {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{ - 4} \over { - 9}}} \right) = {{{4^{1 - {1 \over 3}}}} \over {{3^{2 - {1 \over 3}}}}} = {{{4^{2/3}}} \over {{3^{5/3}}}}$$
$$ = {{{{(8 \times 2)}^{1/3}}} \over {{{(27 \times 9)}^{1/3}}}} = {2 \over 3}{\left( {{2 \over 9}} \right)^{1/3}}$$
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