Right lim < x → 0+>; (2^x- 1)/[2+2^( 1/x)]= 0，

Because the molecular limit is 0 and the denominator limit is infinite.

Then lim

6. Left straight line and left straight line

Right lim < x → 0+>; {[2+e^( 1/x)]/[ 1+e^(4/x)]+sinx/x}

= lim & ltx→0+>{[2e^(-4/x)+e^(-3/x)]/[e^(-4/x)+ 1]+sinx/x}

=(0+0)/(0+ 1)+ 1 = 1.

Then lim