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