# MATH V1101 Study Guide - Midterm Guide: Quotient Rule, Power Rule, 32X

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Math V1101: Calculus I (Chao Li) Midterm I (Solutions to sample and practice problems)

Sample A

Problem 1. (8 points) Mark True or False. No justiﬁcation is needed.

FFTF FTFF

Problem 2.

b, a, c, d, e

Problem 3.

b, e, a, d, c

Problem 4.

(1) lim

x→1

x2−4x+ 3

x−1= lim

x→1

(x−3)(x−1)

x−1= lim

x→1(x−3) = −2.

(2) lim

x→1

x2−1

|x−1|does not exist. In fact,

x2−1

|x−1|=

x2−1

x−1=x+ 1, x > 1,

x2−1

−(x−1) =−(x+ 1), x < 1.

So

lim

x→1+

x2−1

|x−1|= lim

x→1+(x+ 1) = 2,

and

lim

x→1−

x2−1

|x−1|= lim

x→1−−(x+ 1) = −2.

The two one-sided limits exist but they are not the same, therefore lim

x→1

x2−1

|x−1|does not

exist.

(3) lim

x→+∞

√x2+ 4 −x= lim

x→+∞

4

√x2+ 4 + x= 0.

Problem 5. (Derivatives)

(1) f(x) = x2(1 −2√x) = x2−2x5/2, using the power rule we get

f′(x) = 2x−2(5/2·x3/2) = 2x−5x3/2.

(2) f(x) = x+ 3

x2+ 1. Using the quotient rule we get

f′(x) = (x+ 3)′(x2+ 1) −(x+ 3)(x2+ 1)′

(x2+ 1)2=x2+ 1 −(x+ 3) ·2x

(x2+ 1)2=−x2−6x+ 1

(x2+ 1)2.

1

## Document Summary

Problem 1. (8 points) mark true or false. Problem 4. (1) lim x 1 (2) lim x 1. = lim x 1 x 1 does not exist. In fact, x2 4x + 3 x 1 x2 1. = lim x 1 (x 3) = 2. x2 1. = (x + 1), x < 1. x2 1 x 1 x2 1. = lim x 1+ (x + 1) = 2, lim x 1 . = lim x 1 (x + 1) = 2. The two one-sided limits exist but they are not the same, therefore lim x 1 exist. (3) lim x + . X2 + 4 x = lim x + . Using the quotient rule we get x2 1. |x 1| does not f (x) = (x + 3) (x2 + 1) (x + 3)(x2 + 1) (x2 + 1)2 x2 + 1 (x + 3) 2x (x2 + 1)2.