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Calculus 1, Recitation 1

- Midterm 1 Discussion -

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Math-210 Recitation 1 Calculus 1

1 The Limit

limx→c

f(x) = L ⇐⇒ limx→c−

f(x) = limx→c+

f(x) = L

2 Computing Limits Algebraically

1. Direct Subsitution: plug-in

2. Algebraic Manipulation: simplify via factoring

3. Rationalization: multiplying by the conjugate

2.1 Computing Limits Involving as x→ ±∞

Supppose:

f(x) =p(x)

q(x)=

anxn + an−1x

n−1 + · · ·+ a0

bmxm + bm−1xm−1 + · · ·+ b0

Then:

1. If n > m limx→±∞

f(x) =∞ or −∞

2. If n < m limx→+∞

f(x) = limx→−∞

f(x) = 0

3. If n = m limx→±∞

f(x) =anbm

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Math-210 Recitation 1 Calculus 1

3 Continuity Test for Functions

A function y = f(x) is continuous at a point x = c if it meets the following three conditions:

1. f(c) is defined

2. limx→c

f(x) exists

3. limx→c

f(x) = f(c)

If you go through these checkpoints and they’re all true, then y = f(x) is continuous atx = c and if it’s not, then y = f(x) is not continuous at x = c. It’s that simple.

4 The Formal Definition of the Derivative

The derivative of f(x) with respect to x is defined as:

f ′(x) = lim∆x→0

f(x+ ∆x)− f(x)

∆x

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Math-210 Recitation 1 Calculus 1

Problem 1

limx→1

x− 1√x2 + 3− 2

= limx→1

x− 1√x2 + 3− 2

·√x2 + 3 + 2√x2 + 3 + 2

= limx→1

(x− 1)(√x2 + 3 + 2)

(x2 + 3)− 4

= limx→1

(x− 1)(√x2 + 3 + 2)

x2 − 1

= limx→1

(x− 1)(√x2 + 3 + 2)

(x− 1)(x+ 1)

= limx→1

√x2 + 3 + 2

x+ 1

= limx→1

√(1)2 + 3 + 2

(1) + 1

= limx→1

√4 + 2

2

= 2

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Math-210 Recitation 1 Calculus 1

Problem 2A

limx→−∞

2x− x2

3x+ 5= lim

x→−∞

2x− x2

3x+ 5· 1/x

1/x

= limx→−∞

2− x

3 +5

x

= ∞

Problem 2B

limx→∞

3x2 + x− 7

2x5 − 6x4 + 1= lim

x→∞

3x2 + x− 7

2x5 − 6x4 + 1· 1/x2

1/x2

= limx→∞

3 +1

x− 7

x2

2x3 − 6x2 +1

x2

= 0

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Math-210 Recitation 1 Calculus 1

Problem 3

Check the following function for continuity at x = −3, 3.

f(x) =

x3 − 27

x2 − 9, if x 6= 3

9

2, if x = 3

limx→3

x3 − 27

x2 − 9= lim

x→3

(x− 3)(x2 + 3x+ 9)

(x− 3)(x+ 3)

= limx→3

x2 + 3x+ 9

x+ 3

=(3)2 + 3(3) + 9

(3) + 3

=27

6

=9

2

f(3) =9

2

• limx→3

f(x) = f(3) =9

2, therefore f is continuous at x = 3

• Notice at x = −3, f is not defined (division by zero) therefore

f is NOT continuous at x = −3

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Math-210 Recitation 1 Calculus 1

Problem 4

Use the formal definition of the derivative to find the derivative of f(x) = 5x2 at x = −1.

f ′(x) = limh→0

f(x+ h)− f(x)

h= lim

h→0

5(x+ h)2 − 5x2

h

= limh→0

5(x2 + 2xh+ h2)− 5x2

h

= limh→0

5x2 + 10xh+ 5h2 − 5x2

h

= limh→0

5x2 + 10xh+ 5h2 − 5x2

h

= limh→0

10xh+ 5h2

h

= limh→0

h(10x+ 5h)

h

= limh→0

(10x+ 5h)

= limh→0

(10x+ 5h)

= 10x

f ′(x) = 10x

f ′(−1) =dy

dx

∣∣∣x=−1

= −10

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Math-210 Recitation 1 Calculus 1

Problem 5A

Find f (4)

f(x) = −3 ln

[sin(π) + e3/2

sec(π/3) tan(π/3)

]−2/3

f ′(x) = 0

f ′′(x) = 0

f (3)(x) = 0

f (4)(x) = 0

Problem 5B

A function f(x) = k, where k is a constant has a derivative at x = xi of . Provide agraphical representation to accompany your answer.

x

y

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Math-210 Recitation 1 Calculus 1

Problem 6

Differentiate: y = 2cotx

y′ = 2cotx ln 2 · d

dx

[cotx

]= 2cotx ln 2 ·

[− csc2 x

]= −(ln 2) 2cotx csc2 x

Problem 7

Differentiate: g(x) = 4 ln(ln(ln(secx)))

g′(x) =4

ln (ln(secx))· d

dx

[ln (ln(secx))

]

=4

ln (ln(secx))· 1

ln(secx)· d

dx

[ln(secx)

]

=4

ln (ln(secx))· 1

ln(secx)· 1

secx· d

dx

[secx

]

=4

ln (ln(secx))· 1

ln(secx)· 1

secx·[

secx tanx]

=4 secx tanx

ln(ln(secx)) ln(secx) secx

=4 tanx

ln(ln (secx)) ln(secx)

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Math-210 Recitation 1 Calculus 1

Problem 8

Finddy

dx

∣∣∣(0,0)

of exy = e4x − e5y.

exy = e4x − e5y

d

dx

[exy]

=d

dx

[e4x − e5y

]

exy · d

dx

[xy]

= 4e4x − 5e5y y′

exy ·[xy′ + y

]= 4e4x − 5e5y y′

xexyy′ + yexy = 4e4x − 5e5y y′

xexyy′ + 5e5y y′ = 4e4x − yexy

y′[xexy + 5e5y

]= 4e4x − yexy

dy

dx=

4e4x − yexy

xexy + 5e5y

dy

dx

∣∣∣(0,0)

=4e4(0) − (0)e0

(0)e0 + 5e5(0)=

4

5

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Math-210 Recitation 1 Calculus 1

Problem 9A

The following function represents the derivative of a function f at the point (a, f(a)).Identify f and a. (You should also remember that ∆x = h.)

lim∆x→0

sin(π

6+ ∆x

)− sin

(π6

)∆x

f(x) = sin x, a =π

6

Problem 9B

Given f is continuous and differentiable on [a, b] and a ≤ c ≤ b. Compare and contrast thefollowing:

lim∆x→0

f(x+ ∆x)− f(x)

∆xand lim

∆x→0

f(c+ ∆x)− f(c)

∆x

The first limit is equivalent to f ′(x) and will produce a derivative as a function of x,tracing the slope across [a, b]. The second limit is equivalent to f ′(c) which will computethe derivative at x = c for the function f . In other words the first limit will produce afunction and the second limit will produce a numerical value.

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Math-210 Recitation 1 Calculus 1

Problem 10

Find the derivative of : y = 3t cos

(t

3

)

y′ = 3t · ddt

[cos

(t

3

)]+ cos

(t

3

)· ddt

(3t)

= 3t

[− sin

(t

3

)]· ddt

[t

3

]+ 3 cos

(t

3

)

= 3t

[− sin

(t

3

)](1

3

)+ 3 cos

(t

3

)

= t cos

(t

3

)− t sin

(t

3

)

For Fun

Try differentiating

f(x) =

√3x+ 2

2x− 1

Hopefully you get

−7

2(3x+ 1)1/2 (2x− 1)3/2

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