MAT137Y1 Midterm: 2010 Test 1 solution

• School: UTSG
• Department: Mathematics
• Course: MAT137Y1
• Professor: N/ A

Document Summary

Mat137y, 2010-2011 winter session, solutions to term test 1: evaluate the limits which exist. (do not prove them using the formal de nition of limit. You may not use l"hopital"s rule. ) tan(3x2) + sin2(5x) (a) (8%) lim x 0. = lim x 0 lim x 0 and lim x 0 sin(3x2) 3 (cid:18)3 (cid:19) (cid:19)2 25 = (12)(25) = 25. cos(3x2) Since x > 3 in the limit, we see that |x 3| = x 3, so. |x 3| = lim x 3+ x 3 x 3. = lim x 3+ lim x 3+ which does not exist. (c) (8%) lim x 1 x x2 x 1 x2 x 1 lim x 1. Multiplying numerator and denominator by x2 + x, we see that x4 x (x 1)(x2 + = lim x 1 x) x) (x2 x) (x 1) (x2 + (x2 + x(x3 1) (x 1)(x2 + x) x(x2 + x + 1)