5 - x ind lim 2-5 1 - Vx -4 plution: The first step is ALWAYS to plug in a value: 5 - 20 5- 5 m 5 1 - Vx - 4 1 - VI so this limit is the - indeterminate form. find the limit, we multiply the numerator and denominator by the conjugate of the xpression containing a square root: 5 - 20 (5 - 20) (1 + Vac - 4) m = lim = lim (5 - x) (1+ vx - 5 1 - c - 4 x 5 (1 - vx - 4) (1 + vx - 4) x-5 12 - ( Vx - 4)2 5 - 2) (1+ x -4 (5 -2 1 + Vx-4 lim lim = lim (1+ Vx - 4) x-+5 1 - (2 - 4) x-+5 c-+5 O 1/2 0 0 0 -1 0 1 0 2. . a: 5 Flnd 11m . 3255* m2 10.1: + 25 Solution: The rst step is ALWAYS to plug in a value: a: 5 5 5 0 gig]: m : m : 650 this limit is the 6 indeterminate form. We use a2 2ab + b2 = (a b)2to notice that m2 103: + 25 = (a: 5)2.Thus we write x 5 . :1: 5 1 33?5 m2 103: + 25 9H5- (a: 5)2 :1:>5_ a: 5 However, if we plug in the value, we get 1. 1, 1 1 1 1m 2 1m 2 = . ac>5 m 5 5 5 0 non-zero . In such cases, , we must analyze HOW the denominator approaches 0. Since we approach 5 from the left, we approach 5 being smaller than 5. Thus we can plug in, say, 4.9: a: 5 : 4.9 5 : 0.1 5_ (I) 5 O O DNE OOO *OO OO