#lim_(x->4) (sqrt(1+2x)-3)/(sqrtx-2)# ?

1 Answer
Apr 3, 2018

#lim_(x->4) (sqrt(1+2x)-3)/(sqrtx-2) = 4/3#

Explanation:

#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#