Question 18 of 401MediumMCQ
x→0+lim(tan−13x)2(e5(x)34−1)tan(5(x)31)loge(1+3x2) is equal to
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Official solution
limx→0+(tan−1(3x))2(e5x34−1)tan(5(x)31)ln(1+3x3)
\[\mathop {\lim }\limits_{x \to {0^ + }} {{{{\tan \left( {5{{(x)}^{{1 \over 3}}}} \right)} \over {5{{(x)}^{{1 \over 3}}}}}{{\ln (1 + 3{x^2})} \over {3{x^2}}} \times 5{{(x)}^{{1 \over 3}}}(3{x^2})} \over {{{{{\left( {{{\tan }^2}\left( {3\sqrt x } \right)} \right)}^2}} \over {{{\left( {3\sqrt x } \right)}^2}}}{{\left( {{e^{5x{4 \over 3}}} - 1} \right)} \over {5{x^{{1 \over 3}}}}} \times 9x \times 5{x^{{4 \over 3}}}}}\]
\[\mathop {\lim }\limits_{x \to {0^ + }} {{{{\tan \left( {5{{(x)}^{{1 \over 3}}}} \right)} \over {5{{(x)}^{{1 \over 3}}}}}{{\ln (1 + 3{x^3})} \over {3{x^2}}} \times 15{x^{{7 \over 3}}}} \over {{{{{\left( {{{\tan }^2}\left( {3\sqrt x } \right)} \right)}^2}} \over {{{\left( {3\sqrt x } \right)}^2}}}{{\left( {{e^{5(x){4 \over 3}}} - 1} \right)} \over {5{x^{{1 \over 3}}}}} \times 45{x^{{7 \over 3}}}}}\]
\[ = {1 \over 3}\]
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