If f(x)= (e^ x-2 -1 ) (x-1) , then _ x 2 f(x) is equal to

Step-by-step solution

aligned & f(x)= (e^ x-2 -1 ) (x-1) & _ x 2 (e^ x-2 -1 ) (x-1) =L aligned It is 0 0 (undefined) condition so using L'hospital's rule aligned & L = _ x 2 [ ( e ^ x -2 -1 ) ⁻ ( x -1) ⁻ ] & L = _ x 2 ( e ^ x -2 -1 ) e ^ ( x -2) 1 /( x -1) & L = _ x 2 ( e ²⁻²-1 ) e ²⁻² (2-1) & L = (0) e ^0 1 & ~L =1 aligned