using direct substitution to evaluate the limit

Question

use direct substitution to show that each limit leads to the indeterminate form 0/0. Then, evaluate the limit using algebra.

1. lim √ x+4 −1/ x+3

 x→−3

2, lim 2x² +7x−4x/ x²+x−2

x→1+

Answer 1

(sqr(x+4) -1)/(x+3)

(sqr(-3+4)-1)/(-3+3) = (sqr1)-1)/0 = (1-1)/0=0/0

take derivatives separately of denominator and numerator, apply L'Hopital's rule

(1/2)/sqr(x+4))/1 = 1/2sqr(x+4) = 1/2sqr(-3+4)= 1/2(1) = 1/2 = limit as x approaches -3

UNLESS you meant x approaches 3 from the left side, but that seems unlikely as then (sqr7-1)/6 doesn't =0/0 or any indeterminate form

lim (2x^2 -7x +4)/(x^2+x-2) as x approaches 1+

(2-7+4)/(1+1-2)= 1/0 indeterminate

take derivatives to get

(4x -7)/(2x +1) = (4-7)/(2+1) = -3/3 = -1 = limit as x approaches 1