APPM 1350 Midterm: examf_f15_1350_solutions
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Fall 2015: evaluate the following limits. (a) (6 pts) lim. /2 sin(3 ) (c) (6 pts) lim x 0|x| cos(1/x) (b) (6 pts) lim x /4. 1 tan x sin x cos x (d) (8 pts) lim x 0 (1 2x)1/x. Solution: (a) lim x /2 (b) lim x /4 sin(3 ) sin(3 /2) Sec2 x cos x + sin x. 1 tan x sin x cos x. 1 sin x sin x cos x. = lim x /4 cos x sin x cos x(sin x cos x) |x| |x| cos(cid:0) 1 x(cid:1) 1 x(cid:1) |x| Since lim x 0 |x| = 0 and lim x 0|x| = 0, by the squeeze theorem, lim x 0|x| cos(1/x) = 0 . (d) let. L = lim x 0 ln l = lim x 0. = lim x 0 (1 2x)1/x. ln(1 2x)1/x ln(1 2x) x. L = e 2: (16 pts, 8 pts each) evaluate the following integrals.
