Author Archives: Engineering Math

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 4

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PROBLEM:

Evaluate lim⁡x→∞(x3+x+2x2−1)\displaystyle \lim\limits_{x\to \infty }\left(\frac{x^3+x+2}{x^2-1}\right)

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Solution:

Divide by the highest denominator power

lim⁡x→∞(x3+x+2x2−1)=lim⁡x→∞(x3+x+2x2−1⋅1x31x3)=lim⁡x→∞(x3x3+xx3+2x3x2x3−1x3)=lim⁡x→∞(1+1x2+2x31x−1x3)=1+0+00−0=∞  (Answer)\begin{align*} \lim\limits_{x\to \infty }\left(\displaystyle \frac{x^3+x+2}{x^2-1}\right) & =\lim\limits_{x\to \infty }\left(\displaystyle \frac{x^3+x+2}{x^2-1}\cdot \displaystyle \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right) \\ \\ &=\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{x^3}{x^3}+\displaystyle \frac{x}{x^3}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{x^2}{x^3}-\displaystyle \frac{1}{x^3}}\right)\\ \\ &=\lim\limits_{x\to \infty }\left(\displaystyle \frac{1+\displaystyle \frac{1}{x^2}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{1}{x}-\displaystyle \frac{1}{x^3}}\right)\\ \\ &=\displaystyle \frac{1+0+0}{0-0}\\ \\ &=\infty \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 3

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PROBLEM:

Evaluate lim⁡x→∞(4x+5x2+1)\displaystyle \lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\right)

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Solution:

Divide by the highest denominator power

lim⁡x→∞(4x+5x2+1)=lim⁡x→∞(4x+5x2+1⋅1x21x2)=lim⁡x→∞(4xx2+5x2x2x2+1x2)=lim⁡x→∞(4x+5x21+1x2)=0+01+0=0  (Answer)\begin{align*} \lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\right) & =\lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\cdot \frac{\displaystyle\frac{1}{x^2}}{\displaystyle\frac{1}{x^2}}\right) \\ \\ &=\lim\limits_{x\to \infty }\left(\frac{\displaystyle\frac{4x}{x^2}+\displaystyle\frac{5}{x^2}}{\displaystyle\frac{x^2}{x^2}+\displaystyle\frac{1}{x^2}}\right)\\ \\ &=\lim\limits_{x\to \infty }\left(\frac{\displaystyle\frac{4}{x}+\displaystyle\frac{5}{x^2}}{1+\displaystyle\frac{1}{x^2}}\right) \\ \\ &=\displaystyle\frac{0+0}{1+0} \\ \\ &=0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 2

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PROBLEM:

Evaluate lim⁡x→∞(3x2+x+2x3+8x+1)\displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\right)

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Solution:

Divide by the highest denominator power.

lim⁡x→∞(3x2+x+2x3+8x+1)=lim⁡x→∞(3x2+x+2x3+8x+1⋅1x31x3)=lim⁡x→∞(3x2x3+xx3+2x3x3x3+8xx3+1x3)=lim⁡x→∞(3x+1x2+2x31+8x2+1x3)=0+0+01+0+0=0  (Answer)\begin{align*} \displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\right) & =\displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\cdot \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right)\\ \\ &=\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{3x^2}{x^3}+\displaystyle \frac{x}{x^3}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{x^3}{x^3}+\frac{8x}{x^3}+\frac{1}{x^3}}\right)\\ \\ & =\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{3}{x}+\frac{1}{x^2}+\displaystyle \frac{2}{x^3}}{1+\displaystyle \frac{8}{x^2}+\frac{1}{x^3}}\right)\\ \\ &=\frac{0+0+0}{1+0+0} \\ \\ &=0\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 1

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PROBLEM:

Evaluate lim⁡x→∞(6x3+4x2+58x3+7x−3) \displaystyle \lim\limits_{x\to \infty }\left( \frac{6x^3+4x^2+5}{8x^3+7x-3}\right)

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Solution:

Divide by the highest denominator power.

lim⁡x→∞(6x3+4x2+58x3+7x−3)=lim⁡x→∞(6x3+4x2+58x3+7x−3⋅1x31x3)=lim⁡x→∞ (6+4x+5x38+7x2−3x3)=lim⁡x→∞ (6+4x+5x3)lim⁡x→∞ (8+7x2−3x3)=6+0+08+0−0=68=34  (Answer)\begin{align*} \displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\right)&=\displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\cdot \displaystyle \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right) \\ \\ & =\displaystyle \lim\limits_{x\to \infty \:}\left(\displaystyle \frac{6+\displaystyle \frac{4}{x}+\displaystyle \frac{5}{x^3}}{8+\displaystyle \frac{7}{x^2}-\frac{3}{x^3}}\right)\\ \\ & =\displaystyle \frac{\lim\limits_{x\to \infty \:}\left(6+\displaystyle \frac{4}{x}+\frac{5}{x^3}\right)}{\lim\limits_{x\to \infty \:}\left(8+\displaystyle\frac{7}{x^2}-\displaystyle \frac{3}{x^3}\right)}\\ \\ & =\displaystyle \frac{6+0+0}{8+0-0}\\ \\ & =\displaystyle \frac{6}{8}\\ \\ & =\displaystyle \frac{3}{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 22

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PROBLEM:

If f(x)=x2−2x+3 \displaystyle f\left(x\right)=x^2-2x+3, find lim⁡x→0(f(x+2)−f(2)x)\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(x+2\right)-f\left(2\right)}{x}\right).

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SOLUTION:

lim⁡x→0(f(x+2)−f(2)x)=lim⁡x→0(((x+2)2−2(x+2)+3)−(22−2⋅2+3)x)=lim⁡x→0(((x+2)2−2(x+2)+3)−(3)x)=lim⁡x→0(((x+2)2−2(x+2))x)\begin{align*} \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(x+2\right)-f\left(2\right)}{x}\right)&=\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(2^2-2\cdot 2+3\right)}{x}\right)\\ \\ & =\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(3\right)}{x}\right)\\ \\ &=\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)\right)}{x}\right)\\ \\ \end{align*}

Direct substitution of x=0x=0 gives the indeterminate form 00\frac{0}{0}. Therefore, we proceed by factoring the numerator.

=lim⁡x→0(x+2)(x+2−2)x=lim⁡x→0(x+2)(x)x=lim⁡x→0(x+2)=0+2=2  (Answer)\begin{align*} & =\displaystyle \lim\limits_{x\to 0}\displaystyle \frac{\left(x+2\right)\left(x+2-2\right)}{x}\\ \\ & =\lim\limits_{x\to 0}\frac{\left(x+2\right)\left(x\right)}{x} \\ \\ & =\lim\limits_{x\to 0}\left(x+2\right) \\ \\ & =0+2 \\ \\ & =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 21

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PROBLEM:

If f(x)=x2−2x+3\displaystyle f\left(x\right)=x^2-2x+3, find lim⁡x→2(f(x)−f(2)x−2)\displaystyle \lim\limits_{x\to 2}\left(\frac{f\left(x\right)-f\left(2\right)}{x-2}\right).

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SOLUTION:

lim⁡x→2(f(x)−f(2)x−2)=lim⁡x→2((x2−2x+3)−(22−2⋅2+3)x−2)=lim⁡x→2((x2−2x+3)−3x−2)=lim⁡x→2(x2−2xx−2)\begin{align*} \displaystyle \lim\limits_{x\to 2}\left(\displaystyle \frac{f\left(x\right)-f\left(2\right)}{x-2}\right) & =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-\left(2^2-2\cdot 2+3\right)}{x-2}\right) \\ \\ & =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-3}{x-2}\right)\\ \\ & =\lim\limits_{x\to 2}\left(\displaystyle \frac{x^2-2x}{x-2}\right)\\ \\ \end{align*}

Direct substitution of x=2x=2 gives the indeterminate form 00\frac{0}{0}. Therefore, we proceed by factoring the numerator.

=lim⁡x→2(x(x−2)x−2)=lim⁡x→2(x)=2  (Answer)\begin{align*} & =\lim\limits_{x\to 2}\left(\displaystyle \frac{x\left(x-2\right)}{x-2}\right) \\ \\ & =\lim\limits_{x\to 2}\left(x\right) \\ \\ & =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 20

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PROBLEM:

If f(x)=x \displaystyle f\left(x\right)=\sqrt{x}, find lim⁡x→0(f(9+x)−f(9)x)\displaystyle \lim\limits_{x\to 0}\left(\frac{f\left(9+x\right)-f\left(9\right)}{x}\right).

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SOLUTION:

lim⁡x→0(f(9+x)−f(9)x)=lim⁡x→0(9+x−9x)\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(9+x\right)-f\left(9\right)}{x}\right)=\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-\sqrt{9}}{x}\right)

Direct substitution of x=0x=0 gives the indeterminate form 00\frac{0}{0}. Therefore, we proceed by rationalizing the numerator.

=lim⁡x→0(9+x−3x⋅9+x+39+x+3)=lim⁡x→0(9+x−9x(9+x+3))=lim⁡x→0(xx(9+x+3))=lim⁡x→0(1(9+x+3))=(1(9+0+3))=16  (Answer)\begin{align*} & =\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-3}{x}\cdot \displaystyle \frac{\sqrt{9+x}+3}{\sqrt{9+x}+3}\right)\\ \\ & =\lim\limits_{x\to 0}\left(\displaystyle \frac{9+x-9}{x\left(\sqrt{9+x}+3\right)}\right)\\ \\ & =\lim\limits_{x\to 0}\left(\displaystyle \frac{x}{x\left(\sqrt{9+x}+3\right)}\right)\\ \\ & =\lim\limits_{x\to 0}\left(\displaystyle \frac{1}{\left(\sqrt{9+x}+3\right)}\right)\\ \\ & =\left(\displaystyle \frac{1}{\left(\sqrt{9+0}+3\right)}\right)\\ \\ & =\displaystyle \frac{1}{6} \ \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 19

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PROBLEM:

If f(x)=xf\left(x\right)=\sqrt{x}, find lim⁡x→4(f(x)−f(4)x−4)\displaystyle \lim\limits_{x\to 4}\left(\frac{f\left(x\right)-f\left(4\right)}{x-4}\right)

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SOLUTION:

lim⁡x→4(f(x)−f(4)x−4)=lim⁡x→4(x−4x−4)\displaystyle \lim\limits_{x\to 4}\left(\displaystyle \frac{f\left(x\right)-f\left(4\right)}{x-4}\right)=\lim\limits_{x\to 4}\left(\displaystyle \frac{\sqrt{x}-\sqrt{4}}{x-4}\right)

Direct substitution of x=4x=4 gives the indeterminate form 00\frac{0}{0}. Therefore, we proceed by rationalizing the numerator.

=lim⁡x→4(x−2x−4)⋅x+2x+2=lim⁡x→4(x−4(x−4)(x+2))=lim⁡x→4(1x+2)=(14+2)=14  (Answer)\begin{align*} & =\lim\limits_{x\to 4}\left(\displaystyle \frac{\sqrt{x}-2}{x-4}\right)\cdot \displaystyle \frac{\sqrt{x}+2}{\sqrt{x}+2} \\ \\ & =\lim\limits_{x\to 4}\left(\displaystyle \frac{x-4}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right) \\ \\ & =\lim\limits_{x\to 4}\left(\displaystyle \frac{1}{\sqrt{x}+2}\right) \\ \\ & =\left(\displaystyle \frac{1}{\sqrt{4}+2}\right) \\ \\ & =\displaystyle \frac{1}{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 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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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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