Write the function as:
#(1+2x)^((2x+5)/x) = (e^ln(1+2x))^((2x+5)/x)= e^((ln(1+2x)(2x+5))/x)#
Now consider the function:
#f(x) = (ln(1+2x)(2x+5))/x = 2ln(1+2x) + 5 ln(1+2x)/x#
we have that:
#lim_(x->0) 2ln(1+2x) = 0#
while:
#lim_(x->0) ln(1+2x)/x #
is in the form #0/0# so we can solve it using l'Hospital's rule:
#lim_(x->0) ln(1+2x)/x = lim_(x->0) (d/dx ln(1+2x))/(d/dx x) = lim_(x->0) 2/(1+2x) = 2#
Putting it togeher:
#lim_(x->0) 2ln(1+2x) + 5 ln(1+2x)/x = 2*0+5*2 = 10#
Now, as #e^x# is a continuous function we have:
#lim_(x->0) e^(f(x)) = e^(lim_(x->0) f(x))#
so that:
#lim_(x->0)(1+2x)^((2x+5)/x) = lim_(x->0) e^((ln(1+2x)(2x+5))/x) =e^10#
