• San Jose State University

• MATH 30

Math 30 - Spring 11 Test 2 Name: Problem 1. Evaluate the limit, if it exists: lim t→0  1 t √ 1 + t − 1 t  . If the limit does not exist, explain why not. Solution: lim t→0  1 t √ 1 + t − 1 t  = lim t→0  1 − √ 1 + t t √ 1 + t  = lim t→0  1 − √ 1 + t t √ 1 + t  ·  1 + √ 1 + t 1 + √ 1 + t  = lim t→0 1 − (1 + t) t √ 1 + t(1 + √ 1 + t) = lim t→0 −t t √ 1 + t(1 + √ 1 + t) = lim t→0 −1 √ 1 + t(1 + √ 1 + t) DS= −1 √ 1(1 + √ 1) = − 1 2 . Problem 2. Evaluate the limit using continuity: lim x→0 e 2x (x 2+9). Explain why this method applies. Solution: f(x) = e 2x (x 2 + 9) is the product of continuous functions and is therefore continuous. This means we can use direct substitution to evaluate the limit: lim x→0 e 2x (x 2 + 9) = e 2·0 (02 + 9) = 1 · 9 = 9