JEE MAIN - Mathematics Hindi (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) के बराबर है :
$$\left( {{2 \over 9}} \right){\left( {{2 \over 3}} \right)^{{1 \over 3}}}$$
$$\left( {{2 \over 3}} \right){\left( {{2 \over 9}} \right)^{{1 \over 3}}}$$
$${\left( {{2 \over 3}} \right)^{{4 \over 3}}}$$
$${\left( {{2 \over 9}} \right)^{{4 \over 3}}}$$
Comments (0)
