Category Archives: Calculus

Includes differential, integral and series calculus

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 11

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Evaluate lim⁡x→3(x−3x−2−4−x) \displaystyle \lim\limits_{x\to 3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\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(x−3x−2−4−x)=lim⁡x→  3(x−3x−2−4−x)⋅x−2+4−xx−2+4−x=lim⁡x→3[(x−3)(x−2+4−x)(x−2−4−x)(x−2+4−x)]=lim⁡x→3[(x−3)(x−2+4−x)2x−6]=lim⁡x→3[(x−3)(x−2+−x+4)2(x−3)]=lim⁡x→3[x−2+4−x2]=3−2+4−32=1  (Answer)\begin{align*} \begin{align*} \lim\limits_{x\to \:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right) & =\lim\limits_{x\to \:\:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)\cdot \frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}} \\ & =\lim\limits_{x\to 3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\left(\sqrt{x-2}-\sqrt{4-x}\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}\right] \\ & =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2x-6}\right]\\ & =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{-x+4}\right)}{2\left(x-3\right)}\right] \\ & =\lim\limits_{x\to 3}\left[\frac{\sqrt{x-2}+\sqrt{4-x}}{2}\right]\\ & =\frac{\sqrt{3-2}+\sqrt{4-3}}{2}\\ & =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*} \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 10

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PROBLEM:

Evaluate lim⁡x→2(x3−8x2−4 )\displaystyle \lim\limits_{x\to 2}\left(\frac{x^3-8}{x^2-4\:}\right)

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SOLUTION:

A straight substitution of  x=2x=2 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→2(x3−8x2−4 )=lim⁡x→2[(x−2)(x2+2x+4)(x+2)(x−2)]=lim⁡x→2[(x2+2x+4)(x+2)]=22+2⋅2+42+2=3  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(\frac{x^3-8}{x^2-4\:}\right) & =\lim\limits_{x\to 2}\left[\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x-2\right)}\right] \\ &=\lim\limits_{x\to 2}\left[\frac{\left(x^2+2x+4\right)}{\left(x+2\right)}\right] \\ & =\frac{2^2+2\cdot 2+4}{2+2} \\ & =3 \ \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 9

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PROBLEM:

Evaluate lim⁡x→4(1x−14x−4 )\displaystyle \lim\limits_{x\to 4}\left(\frac{\frac{1}{x}-\frac{1}{4}}{x-4\:}\right).

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SOLUTION:

A straight substitution of x=4x=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→4(1x−14x−4)=lim⁡x→4(4−x4xx−4)=lim⁡x→44−x4x(x−4)=lim⁡x→4(4−x−4x(4−x))=lim⁡x→4−14x=−14⋅4=−116  (Answer)\begin{align*} \\ \lim\limits_{x\to 4}\left(\frac{\frac{1}{x}-\frac{1}{4}}{x-4}\right)& =\lim\limits_{x\to 4}\left(\frac{\frac{4-x}{4x}}{x-4}\right)\\ \\ & =\lim\limits_{x\to 4}\frac{4-x}{4x\left(x-4\right)}\\ \\ &=\lim\limits_{x\to 4}\left(\frac{4-x}{-4x\left(4-x\right)}\right)\\ \\ & =\lim\limits_{x\to 4}-\frac{1}{4x}\\ \\ & =-\frac{1}{4\cdot 4}\\ \\ & =-\frac{1}{16} \ \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 8

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PROBLEM:

Evaluate lim⁡x→8 x3−2x−8\displaystyle \lim\limits_{x\to 8}\:\frac{\sqrt[3]{x}-2}{x-8}.

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SOLUTION:

A straight substitution of x=8x=8 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→8 x3−2x−8=lim⁡x→ 8 x3−2x−8⋅x23+2x3+4x23+2x3+4=lim⁡x→8 x−8(x−8)(x23+2x3+4)=lim⁡x→8 1(x23+2x3+4)=1(823+283+4)=14+4+4=112  (Answer)\begin{align*} \\ \lim\limits_{x\to 8}\:\frac{\sqrt[3]{x}-2}{x-8}& =\lim\limits_{x\to \:8}\:\frac{\sqrt[3]{x}-2}{x-8}\cdot \frac{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}\\ \\ & =\lim\limits_{x\to 8}\:\frac{x-8}{\left(x-8\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}\\ \\ & =\lim\limits_{x\to 8}\:\frac{1}{\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}\\ \\ & =\frac{1}{\left(\sqrt[3]{8^2}+2\sqrt[3]{8}+4\right)}\\ \\ & =\frac{1}{4+4+4}\\ \\ & =\frac{1}{12} \ \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 7

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PROBLEM:

Evaluate lim⁡x→1 x−1x+3−2 \displaystyle \lim\limits_{x\to 1}\:\frac{x-1}{\sqrt{x+3}-2}.

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 SOLUTION:

A straight substitution of  x=1x=1 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→1 x−1x+3−2=lim⁡x→ 1 x−1x+3−2⋅x+3+2x+3+2=lim⁡x→1(x−1)(x+3+2)(x+3)−22=lim⁡x→1(x−1)(x+3+2)x−1=lim⁡x→1x+3+2=1+3+2=4+2=2+2=4  (Answer)\begin{align*} \\ \lim\limits_{x\to 1}\:\frac{x-1}{\sqrt{x+3}-2}& =\lim\limits_{x\to \:1}\:\frac{x-1}{\sqrt{x+3}-2}\cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}\\ \\ & =\lim\limits_{x\to 1}\frac{\left(x-1\right)\left(\sqrt{x+3}+2\right)}{\left(x+3\right)-2^2}\\ \\ & =\lim\limits_{x\to 1}\frac{\left(x-1\right)\left(\sqrt{x+3}+2\right)}{x-1}\\ \\ & =\lim\limits_{x\to 1}\sqrt{x+3}+2&\\ \\ & =\sqrt{1+3}+2\\ \\ &=\sqrt{4}+2\\ \\ & =2+2\\ \\ & =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 6

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PROBLEM:

Evaluate lim⁡x→0 x+16−4x \displaystyle\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}.

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 SOLUTION:

A straight substitution of x=0x=0 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→0 x+16−4x=lim⁡x→0 x+16−4x⋅x+16+4x+16+4=lim⁡x→0 (x+16)−42x(x+16+4)=lim⁡x→0 x+16−16x(x+16+4)=lim⁡x→0 xx(x+16+4)=lim⁡x→0 1x+16+4= 10+16+4= 14+4= 18  (Answer)\begin{align*} \\ \lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x} & =\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}\cdot \frac{\sqrt{x+16}+4}{\sqrt{x+16}+4}\\ \\ & =\lim\limits_{x\to 0}\:\frac{\left(x+16\right)-4^2}{x\left(\sqrt{x+16}+4\right)}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x+16-16}{x\left(\sqrt{x+16}+4\right)}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x}{x\left(\sqrt{x+16}+4\right)}\\ \\ & =\lim\limits_{x\to 0}\:\frac{1}{\sqrt{x+16}+4}\\ \\ & =\:\frac{1}{\sqrt{0+16}+4}\\ \\ & =\:\frac{1}{4+4}\\ \\ & =\:\frac{1}{8} \ \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 5

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PROBLEM:

Evaluate lim⁡x→0 (x+3)2−92x\displaystyle \lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x}.

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 SOLUTION:

A straight substitution of x=0x=0 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→0 (x+3)2−92x= lim⁡x→0 (x+3)2−(3)22x= lim⁡x→0 (x+3−3)(x+3+3)2x=lim⁡x→0 x(x+6)2x=lim⁡x→0 x+62=0+62=62=3  (Answer)\begin{align*} \lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x} & =\:\lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-\left(3\right)^2}{2x}\\ \\ & =\:\lim\limits_{x\to 0}\:\frac{\left(x+3-3\right)\left(x+3+3\right)}{2x}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x\left(x+6\right)}{2x}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x+6}{2}\\ \\ & =\frac{0+6}{2}\\ \\ & =\frac{6}{2}\\ \\ & =3 \ \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 4

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PROBLEM:

Evaluate lim⁡x→2(x3−x2−x−22x3−5x2+5x−6) \displaystyle \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right).

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 SOLUTION:

A straight substitution of x=2x=2 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→2(x3−x2−x−22x3−5x2+5x−6)=lim⁡x→2((x−2)(x2+x+1)(x−2)(2x2−x+3))=lim⁡x→2(x2+x+12x2−x+3)=22+2+12(2)2−2+3=4+2+18−2+3=79  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right)&=\lim\limits_{x\to 2}\left(\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(2x^2-x+3\right)}\right)\\ \\ & =\lim\limits_{x\to 2}\left(\frac{x^2+x+1}{2x^2-x+3}\right)\\ \\ & =\frac{2^2+2+1}{2\left(2\right)^2-2+3}\\ \\ & =\frac{4+2+1}{8-2+3}\\ \\ & =\frac{7}{9} \ \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 3

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PROBLEM:

Evaluate lim⁡x→3(x3−13x+12x3−14x+15) \displaystyle \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\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(x3−13x+12x3−14x+15)=lim⁡x→3((x−3)(x2+3x−4)(x−3)(x2+3x−5))=lim⁡x→3(x2+3x−4x2+3x−5)=(3)2+3(3)−4(3)2+3(3)−5=9+9−49+9−5=1413  (Answer)\begin{align*} \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\right)& =\lim\limits_{x\to 3}\left(\frac{\left(x-3\right)\left(x^2+3x-4\right)}{\left(x-3\right)\left(x^2+3x-5\right)}\right)\\ \\ & =\lim\limits_{x\to 3}\left(\frac{x^2+3x-4}{x^2+3x-5}\right)\\ \\ &=\frac{\left(3\right)^2+3\left(3\right)-4}{\left(3\right)^2+3\left(3\right)-5}\\ \\ & =\frac{9+9-4}{9+9-5}\\ \\ & =\frac{14}{13} \ \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 2

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PROBLEM:

Evaluate lim⁡x→2(x2+2x−83x−6)\displaystyle \lim\limits_{x\to 2}\left(\frac{x^2+2x-8}{3x-6}\right)

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SOLUTION:

A straight substitution of x=2x=2 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→2(x2+2x−83x−6)=lim⁡x→2((x+4)(x−2)3(x−2))=lim⁡x→2(x+43)=2+43=63=2  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(\frac{x^2+2x-8}{3x-6}\right)& =\lim\limits_{x\to 2}\left(\frac{\left(x+4\right)\left(x-2\right)}{3\left(x-2\right)}\right)\\ \\ &=\lim\limits_{x\to 2}\left(\frac{x+4}{3}\right)\\ \\ &=\frac{2+4}{3}\\ \\ &=\frac{6}{3}\\ \\ & =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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