Category Archives: Calculus

Includes differential, integral and series calculus

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 18

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PROBLEM:

Evaluate lim⁡x→π(sin⁡2(x)1+cos⁡(x)).\displaystyle \lim\limits_{x\to \pi }\left(\frac{\sin^2\left(x\right)}{1+\cos\left(x\right)}\right).

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SOLUTION:

Direct substitution of x=πx=\pi gives the indeterminate form 00\frac{0}{0}. Therefore, we should apply trigonometric identities.

We know the Pythagorean identity, sin⁡2(x)=1−cos⁡2(x)\sin^2\left(x\right)=1-\cos^2\left(x\right). Therefore, we have

lim⁡x→π(sin⁡2(x)1+cos⁡(x))=lim⁡x→π(1−cos⁡2(x)1+cos⁡(x))=lim⁡x→π((1+cos⁡(x))(1−cos⁡(x))1+cos⁡(x))=lim⁡x→π(1−cos⁡(x))=(1−cos⁡(π))=(1−(−1))=2  (Answer)\begin{align*} \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{\sin^2\left(x\right)}{1+\cos\left(x\right)}\right) & = \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{1-\cos^2\left(x\right)}{1+\cos\left(x\right)}\right) \\ \\ & = \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{\left(1+\cos\left(x\right)\right)\left(1-\cos\left(x\right)\right)}{1+\cos\left(x\right)}\right)\\ \\ & =\lim\limits_{x\to \pi }\left(1-\cos\left(x\right)\right)\\ \\ & =\left(1-\cos\left(\pi \right)\right)\\ \\ & =\left(1-\left(-1\right)\right)\\ \\ & =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Mathematics

Algebra
Trigonometry
Geometry
Precalculus
Calculus I
Calculus II
Calculus III
Advanced Engineering Mathematics
Probability and Statistics
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 17

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PROBLEM:

Evaluate lim⁡x→0(sin⁡(x)sin⁡(2x)1−cos⁡(x)).\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right).

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SOLUTION:

Direct substitution of x=0x=0 gives the indeterminate form 00\frac{0}{0}. Therefore, we should apply trigonometric identities.

We know that sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin\left(2x\right)=2\sin\left(x\right)\cos\left(x\right), so we can rewrite the original function as

lim⁡x→0(sin⁡(x)sin⁡(2x)1−cos⁡(x))=lim⁡x→0(sin⁡(x)⋅2(sin⁡(x)cos⁡(x))1−cos⁡(x))=2⋅lim⁡x→0(sin⁡2(x)cos⁡(x)1−cos⁡(x))\begin{align*} \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =\lim\limits_{x\to 0}\left(\frac{\sin\left(x\right)\cdot 2\left(\sin\left(x\right) \cos\left(x\right)\right)}{1-\cos\left(x\right)}\right)\\ \\ & =\displaystyle 2\cdot \lim\limits_{x\to 0}\left(\frac{\sin^2\left(x\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\ \end{align*}

We also know the Pythagorean identity sin⁡2(x)=1−cos⁡2(x)\sin^2\left(x\right)=1-\cos^2\left(x\right). So,

lim⁡x→0(sin⁡(x)sin⁡(2x)1−cos⁡(x))=2⋅lim⁡x→0((1−cos⁡2(x))cos⁡(x)1−cos⁡(x))=2⋅lim⁡x→0((1+cos⁡(x))(1−cos⁡(x))cos⁡(x)1−cos⁡(x))=2⋅lim⁡x→0((1+cos⁡(x))cos⁡(x))=2⋅((1+cos⁡(0))cos⁡(0))=2⋅((1+1)⋅1)=4  (Answer)\begin{align*} \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1-\cos^2\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\ \\ & =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1+\cos\left(x\right)\right)\left(1-\cos\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right) \\ \\ & =2\cdot \lim\limits_{x\to 0}\left(\left(1+\cos\left(x\right)\right)\cos\left(x\right)\right) \\ \\ & = 2\cdot \left(\left(1+\cos\left(0\right)\right)\cos\left(0\right)\right) \\ \\ & =2\cdot \left(\left(1+1\right)\cdot 1\right) \\ \\ & =4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 16

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PROBLEM:

Evaluate lim⁡x→0(1−cos⁡2(x)1+cos⁡(x)) \displaystyle \lim\limits_{x\to 0}\left(\frac{1-\cos^2\left(x\right)}{1+\cos\left(x\right)}\right)

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SOLUTION:

This problem can be solved using a direct substitution of x=0x=0. That is

lim⁡x→0(1−cos⁡2x1+cos⁡x)=1−cos⁡2(0)1+cos⁡(0)=1−11+1=0  (Answer)\begin{align*} \lim\limits_{x\to 0}\left(\frac{1-\cos^2 x }{1+\cos x }\right) & =\frac{1-\cos^2\left(0\right)}{1+\cos\left(0\right)} \\ \\ & =\frac{1-1}{1+1}\\ \\ & = 0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 15

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PROBLEM:

Evaluate lim⁡x→0(sin⁡3xsin⁡x−tan⁡x) \displaystyle \lim_{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right).

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SOLUTION:

A straight substitution of x=π4\displaystyle x=\frac{\pi }{4} leads to the indeterminate form 00\displaystyle \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→0(sin⁡3xsin⁡x−tan⁡x)=lim⁡x→0(sin⁡3xsin⁡x−sin⁡xcos⁡x)=lim⁡x→0(sin⁡3xsin⁡xcos⁡x−sin⁡xcos⁡x)=lim⁡x→0(sin⁡3xcos⁡xsin⁡xcos⁡x−sin⁡x)=lim⁡x→ 0(sin⁡3xcos⁡x(sin⁡x)(cos⁡x−1))=lim⁡x→0(sin⁡2xcos⁡x(cos⁡x−1))=lim⁡x→0((1−cos⁡2x)⋅cos⁡x−(1−cos⁡x))=lim⁡x→0((1+cos⁡x)(1−cos⁡x)⋅cos⁡x−(1−cos⁡x))=lim⁡x→0((1+cos⁡x)⋅cos⁡(x)−1)=−1⋅lim⁡x→0((1+cos⁡x)⋅cos⁡x)=−1⋅(1+cos⁡0)⋅cos⁡0=−1⋅(1+1)⋅1=−2  (Answer)\begin{align*} \displaystyle \lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right) & =\lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\frac{\sin x}{\cos x}}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\sin^3 x}{\frac{\sin x \cos x - \sin x}{\cos x}}\right) \\ \\ & = \lim _{x\to 0}\left(\frac{\sin^3 x \cos x }{\sin x \cos x - \sin x}\right) \\ \\ &=\lim _{x\to \:0}\left(\frac{\sin^3 x \cos x}{\left(\sin x \right)\left(\cos x-1\right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\sin^2 x \cos x }{\left(\cos x -1\right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\left(1-\cos^2 x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\left(1+\cos x \right) \left(1-\cos x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\left(1+\cos x \right) \cdot \cos\left(x\right)}{-1}\right) \\ \\ & =-1\cdot \lim _{x\to 0}\left(\left(1+\cos x\right)\cdot\cos x \right) \\ \\ & =-1\cdot \left(1+\cos 0 \right)\cdot \cos 0 \\ \\ & =-1\cdot \left(1+1\right)\cdot 1 \\ \\ & =-2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 14

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PROBLEM:

Evaluate lim⁡x→π4(tan⁡ 2xsec⁡ 2x)\displaystyle \lim\limits_{x\to \frac{\pi }{4}}\left(\frac{\tan\:2x}{\sec\:2x}\right).

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SOLUTION:

A straight substitution of x=π4x=\frac{\pi }{4} leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→ π 4(tan⁡ 2xsec⁡ 2x)=lim⁡x→π4(sin⁡ 2xcos⁡ 2x1cos⁡ 2x)=lim⁡x→ π 4(sin⁡ 2x)=sin⁡(2⋅π4)=sin⁡π2=1  (Answer)\begin{align*} \displaystyle \lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\frac{\tan\:2x}{\sec\:2x}\right) & =\lim\limits_{x\to \frac{\pi }{4}}\left(\frac{\frac{\sin\:2x}{\cos\:2x}}{\frac{1}{\cos\:2x}}\right) \\ \\ & =\lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\sin\:2x\right) \\ \\ & =\sin\left(2\cdot \frac{\pi }{4}\right) \\ \\ & =\sin\frac{\pi }{2} \\ \\ & =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 13

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PROBLEM:

Evaluate lim⁡x→3(x2−9x−3)\displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\right).

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SOLUTION:

A straight substitution of x=3x=3 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→ 3(x2−9x−3)=lim⁡x→3(x2−9x−3⋅x2−9x2−9)=lim⁡x→3((x+3)(x−3)(x−3)x2−9)=lim⁡x→3(x2−9(x−3)x2−9)=lim⁡x→3(x+3x2−9)=3+332−9=60=∞  (Answer)\begin{align*} \lim\limits_{x\to \:3}\left(\frac{\sqrt{x^2-9}}{x-3}\right) & =\lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\cdot \frac{\sqrt{x^2-9}}{\sqrt{x^2-9}}\right) \\ \\ & =\lim\limits_{x\to 3}\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\sqrt{x^2-9}}\right) \\ \\ & =\lim\limits_{x\to 3}\left(\frac{x^2-9}{\left(x-3\right)\sqrt{x^2-9}}\right) \\ \\ & =\lim _{x\to 3}\left(\frac{x+3}{\sqrt{x^2-9}}\right) \\ \\ & =\frac{3+3}{\sqrt{3^2-9}} \\ \\ & =\frac{6}{0} \\ \\ & =\infty \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Since the function’s limit is different from the left to its limits from the right, the limit does not exist. 

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Cover photo for Chapter 1: Limits of the textbook Differential and Integral Calculus by Feliciano and Uy

Chapter 1: Limits

Exercise 1.1: Functional Notation

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Exercise 1.2: Theorems on Limits

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

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Exercise 1.3: Indeterminate Forms

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

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Exercise 1.4: Limit at Infinity

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

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Exercise 1.5: Continuity

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

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Exercise 1.6: Asymptotes

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Differential and Integral Calculus by Feliciano and Uy Banner

Differential and Integral Calculus by Feliciano and Uy Complete Solution Manual

Chapter 1: Limits
Chapter 2: Differentiation of Algebraic Functions
Chapter 3: Some Applications of the Derivatives
Chapter 4: Differentiation of Transcendental Functions
Chapter 5: The Indeterminate Forms
Chapter 6: The Differential
Chapter 7: Derivatives From Parametric Equations; Radius and Center of Curvature
Chapter 8: Partial Differentiation
Chapter 9: The Indefinite Integral
Chapter 10: Methods of Integration
Chapter 11: The Definite Integral
Chapter 12: Geometric Applications of the Definite Integral
Chapter 13: Physical Applications of the Definite Integral
Chapter 14: Double Integration
Chapter 12: Geometric Applications of the Definite Integral
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 12

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PROBLEM:

Evaluate lim⁡x→0(1x(13−1x+9))\displaystyle \lim\limits_{x\to 0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right)

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SOLUTION: 

A straight substitution of x=0x=0 leads to the indeterminate form 0⋅00\cdot 0 which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→ 0(1x(13−1x+9))=lim⁡x→0(1⋅ (13−1x+9)x)=lim⁡x→0(x+9−33x+9x)=lim⁡x→0(x+9−33xx+9)=lim⁡x→0(x+9−33xx+9)⋅x+9+3x+9+3=lim⁡x→0(x3x(x+9+3)x+9)=lim⁡x→0(13(x+9+3)x+9)=13(0+9+3)0+9=154  (Answer)\begin{align*} \lim\limits_{x\to \:0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right) & =\lim\limits_{x\to 0}\left(\frac{1\cdot \:\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)}{x}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\frac{\sqrt{x+9}-3}{3\sqrt{x+9}}}{x}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right)\cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} \\ \\ & =\lim\limits_{x\to 0}\left(\frac{x}{3x\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{1}{3\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\ & =\frac{1}{3\left(\sqrt{0+9}+3\right)\sqrt{0+9}} \\ \\ & =\frac{1}{54} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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