Q.

limx→0+tan (5(x)13)loge(1+3x2)(tan–13x)2(e5(x)43–1) is equal to           [2025]

1 1  
2 53  
3 13  
4 115  

Ans.

(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.