Calculus- Can someone please help with determining limits?

Question

a) lim x-->2 [sqr(x) - sqr(2)] / [x -2]

b) lim x-->5 [17/(x+12) - 1] / [x-5]

an explanation along with the answer would be greatly appreciated. thanks!

Answer 1

a) lim x-->2 [sqr(x) - sqr(2)] / [x -2]

rationalize the numerator: mult. numerator and denominator by [sqr(x) + sqr(2)]

lim x-->2 [sqr(x) - sqr(2)] / [x -2]

lim x-->2 [sqr(x) - sqr(2)][sqr(x) + sqr(2)] / [x -2][sqr(x) + sqr(2)]

lim x-->2 [x - 2] / [x -2][sqr(x) + sqr(2)]

lim x-->2 1 / [sqr(x) + sqr(2)] = 1 / [sqr(2) + sqr(2)] = 1 / 2sqr(2) = sqr(2) / 4

b) lim x-->5 [17/(x+12) - 1] / [x-5]

rationalize the fraction: mult. numerator and denominator by (x+12)

lim x-->5 [17/(x+12) - 1](x+12) / [x-5](x+12)

lim x-->5 [17 - (x+12)] / [x-5](x+12)

lim x-->5 [5 - x] / [x-5](x+12)

lim x-->5 -1[x - 5] / [x-5](x+12)

lim x-->5 -1 / (x+12) = -1 / 17