limx→0+tan (5(x)13)loge(1+3x2)(tan–13x)2(e5(x)43–1) is equal to [2025]
(3)
We have, limx→0+tan (5(x)13)loge(1+3x2)(tan–1(3x))2(e5(x)43–1)
⇒ limx→0+tan (5(x)13) loge(1+3x2)3x2×5(x)13(3x2)5(x)13(tan–1(3x))2(3x)2(e5x43–1)5x43×9x×5x43
⇒ limx→0+tan (5(x)13)5(x)13loge(1+3x2)3x2×15x73(tan–1(3x))2(3x)2(e5(x)43–1)5x43×45x73=13.