Question

please solve this link

Answer 1

Lim ( x-> 0 ) [ 1 + x ] ⁶ - 1 / [ ( 1 + x ) ² - 1 ]

Let ( 1 + x ) be y.

Then, ( 1 + x ) = y, x --> y , y --> 0.

Lim ( y -> 0 ) [ y] ⁶ - 1 / { [ y ] ² - 1 }

Lim ( y -> 0) ( y ⁶ - 1 / y - 1 ) / ( y² - 1 / y - 1 )

=> 6 * 1 / 2 * 1

=> 6 / 2 = 3.