limit question (urgent)

Note that: loga(x)^b = b logax
then v ln(u) = v logeu = loge(u)^v


107.
Let u = 1 + (x/2) and v = 2x

limn→∞ 2x ln[1 + (2/x)]
= limn→∞­ ­v ln(u)
= limn→∞­ ln(u)^v
= limn→∞ ln[1 + (2/x)]^2x
= limn→∞ ln{[1 + (2/x)]^x}^2
= ln{ limn→∞ [1 + (2/x)]^x}^2
= ln(e^2)^2
= logee^4
= 4 logee
= 4


108.
Let u = 2x + 1 and v = 1/3x

limn→0 ln(2x + 1) / 3x
= limn→0 (1/3x) ln(2x + 1)
= limn→0 v ln u
= limn→0 ln(u)^v
= limn→0 ln(2x + 1)^(1/3x)
= limn→0 ln(y + 1)^(2/3y) .... [Take y = 2x]
= ln[limn→0 (1 + y)^(1/y)]^(2/3)
= ln(e)^(2/3)
= (2/3)ln(e)
= (2/3)logee
= 2/3