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Math 1101 Calculus I Practice Midterm 1 Solutions 2. Compute the following limits, if they exist. If the limit does not exist, explain why. (a) (3 points) lim x!3 x 2 x2 5x+ 6 Solution: lim x!3 x 2 x2 5x+ 6 = lim x!3 x 2 (x 2)(x 3) == limx6=3 x!3 1 (x 3) Note 1 (x 3) goes to in nity at x = 3 and thus the limit does not exist. To
Page 1 of 159. A Collection of Problems in Differential Calculus Problems Given At the Math 151 - Calculus I and Math 150 - Calculus I With Review Final Examinations Department of Mathematics, Simon Fraser University 2000 - 2010 Veselin Jungic · Petra Menz · Randall Pyke Department Of Mathematics Simon Fraser University
201-103-RE - Calculus 1 WORKSHEET: LIMITS 1. Use the graph of the function f(x) to answer each question. Use 1, 1 or DNEwhere appropriate. (a) f(0) = (b) f(2) = (c) f(3) = (d) lim x!0 f(x) = (e) lim x!0 f(x) = (f) lim x!3+ f(x) = (g) lim x!3 f(x) = (h) lim x!1 f(x) = 2. Use the graph of the function f(x) to answer each question. Use 1, 1 or ...
Unit 1 Practice Test: Limits. Date: Here’s your chance to show what you know! You’ve got all you need in your brain, so trust yourself and put your calculator away. Make sure you show me all the cool work you can do to get your answer when appropriate. 1. Find the following limits if.
The printed solution that immediately follows a problem statement gives you all the details of one way to solve the problem. You might wish to delay consulting that solution until you have outlined an attack in your
Calculus 1 - Limits - Worksheet 9 – Using the Limit Laws 1. Evaluate this limit using the Limit Laws. Show each step. lim 𝑥→5 (2𝑥2−3𝑥+4) 2. Evaluate this limit using the Limit Laws. Show each step. lim 𝑥→−2 (𝑥3+2𝑥2−1 5−3𝑥)
SOLUTIONS: ONE-SIDED AND TWO-SIDED LIMIT PROBLEMS. 1. Evaluate the one-sided limits below. a) i) lim |x − 2|. x→2−. ii) lim |x − 2|. x→2+. e further obtain x − 2 < 0 by subtracting 2 from both sides of the inequ. lity. The absolute value |x − 2| is therefore equal to −(x − 2) f. g the limit yields lim |x − 2| = lim(2 − x) = 2 �. x→2− x→2−.
