#lim_(x->4) (sqrt(1+2x)-3)/(sqrtx-2)# evaluated at #x = 4# yields the indeterminant form #0/0#, therefore, one should use L'Hôpital's rule.
To use L'Hôpital's rule, one differentiates the numerator and the denominator with respect to the dependent variable:
#lim_(x->4) (sqrt(1+2x)-3)/(sqrtx-2) = lim_(x->4) ((d(sqrt(1+2x)-3))/dx)/((d(sqrtx-2))/dx)#
Compute the derivatives:
#lim_(x->4) (sqrt(1+2x)-3)/(sqrtx-2) = lim_(x->4) (1/(sqrt(1+2x)))/(1/(2sqrtx))#
Simplify:
#lim_(x->4) (sqrt(1+2x)-3)/(sqrtx-2) = lim_(x->4) (2sqrtx)/(sqrt(1+2x))#
The limit on the right can be evaluated at #x = 4#
#lim_(x->4) (sqrt(1+2x)-3)/(sqrtx-2) = (2sqrt4)/(sqrt(1+2(4)))#
Simplify:
#lim_(x->4) (sqrt(1+2x)-3)/(sqrtx-2) = 4/3#
