_ x 2 x^2+2^ 2-x -5 2^ -x / 2 -2^ 1-x is equal to _____

Step-by-step solution

We have, _ x 2 2^x+2^ 2-x -5 2^ -x / 2 -2^ 1-x [ 0 0 form ] So, apply L' Hospital's rule = _ x 2 d d x (2^x+2^ 2-x -5 ) d d x (2^ -x / 2 -2^ 1-x ) aligned & = _ x 2 2^x 2-2^ 2-x 2 2^ 1-x 2- 2^ - x 2 2 2 & = _ x 2 2^ x+1 -2^ 3-x 2^ 2-x -2^ - x 2 & = 2^3-2^1 2^0-2⁻¹ = 8-2 2- 1 2 = 12 2-1 =12 aligned