Category Archives: Calculus

Includes differential, integral and series calculus

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 1

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PROBLEM:

Evaluate lim⁡x→4(x3−64x2−16)\displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)

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SOLUTION:

A straight substitution of x=4 x=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(x3−64x2−16)=lim⁡x→4((x−4)(x2+4x+16)(x+4)(x−4))=lim⁡x→4(x2+4x+16x+4)=(4)2+4(4)+164+4=488=6  (Answer)\begin{align*} \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)& =\lim\limits_{x\to 4}\left(\frac{\left(x-4\right)\left(x^2+4x+16\right)}{\left(x+4\right)\left(x-4\right)}\right)\\ \\ & =\lim\limits_{x\to 4}\left(\frac{x^2+4x+16}{x+4}\right)\\ \\ & =\frac{\left(4\right)^2+4\left(4\right)+16}{4+4}\\ \\ & =\frac{48}{8}\\ \\ & =6 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 8

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PROBLEM:

Evaluate lim⁡x→0(3x+2x2−2x+4)\displaystyle \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right).

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SOLUTION:

Plug in the value x=0.

lim⁡x→0(3x+2x2−2x+4)=3(0)+2(0)2−2(0)+4=0+20−0+4=24=12  (Answer)\begin{align*} \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right)& =\frac{3\left(0\right)+2}{\left(0\right)^2-2\left(0\right)+4}\\ \\ & =\frac{0+2}{0-0+4}\\ \\ & =\frac{2}{4}\\ \\ & =\frac{1}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 7

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PROBLEM:

Evaluate lim⁡x→3(3xxx+1)\displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{3x}}{x\sqrt{x+1}}\right).

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SOLUTION:

Plug the value x=3.

lim⁡x→3(3xxx+1)=3(3)33+1=934=33⋅2=36=12  (Answer)\begin{align*} \lim\limits_{x\to 3}\left(\frac{\sqrt{3x}}{x\sqrt{x+1}}\right)&=\frac{\sqrt{3\left(3\right)}}{3\sqrt{3+1}} \\ \\ &=\frac{\sqrt{9}}{3\sqrt{4}} \\ \\ & =\frac{3}{3\cdot 2}\\ \\ & =\frac{3}{6}\\ \\ &=\frac{1}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 6

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PROBLEM:

Evaluate lim⁡x→2(4x−3)(x2+5)\displaystyle \lim_{x\to 2}\left(4x-3\right)\left(x^2+5\right).

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SOLUTION:

Plug the value x=2.

lim⁡x→2(4x−3)(x2+5)=[(4⋅2)−3][(2)2+5]=[8−3][4+5]=(5)(9)=45  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(4x-3\right)\left(x^2+5\right) & =\left[\left(4\cdot 2\right)-3\right]\left[\left(2\right)^2+5\right]\\ & =\left[8-3\right]\left[4+5\right]\\ & =\left(5\right)\left(9\right)\\ & =45 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 5

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PROBLEM:

Evaluate lim⁡x→8(2x+x3−4)\displaystyle \lim\limits_{x\to 8}\left(2x+\sqrt[3]{x}-4\right).

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SOLUTION:

Plug in the value x=8.

lim⁡x→8(2x+x3−4)=[2(8)+83−4]=[16+2−4]=14  (Answer)\begin{align*} \lim\limits_{x\to 8}\left(2x+\sqrt[3]{x}-4\right) & = \left[2\left(8\right)+\sqrt[3]{8}-4\right]\\ & =\left[16+2-4\right]\\ & =14 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 4

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PROBLEM:

Evaluate lim⁡x→π3(sin⁡ 2xsin⁡ x)\displaystyle \lim\limits _{x\to \frac{\pi }{3}}\left(\frac{\sin\:2x}{\sin\:x}\right).

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SOLUTION:

Plug in the value x=π3\displaystyle x=\frac{\pi }{3}.

lim⁡x→π3(sin⁡ 2xsin⁡ x)=sin⁡(2⋅π3)sin⁡ (π3)=3232=1  (Answer)\begin{align*} \lim\limits_{x\to \frac{\pi }{3}}\left(\frac{\sin\:2x}{\sin\:x}\right) & =\frac{\sin\left(2\cdot \frac{\pi }{3}\right)}{\sin\:\left(\frac{\pi }{3}\right)} \\ & =\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\\ & =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 3

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PROBLEM:

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

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SOLUTION:

lim⁡x→π4(tan⁡ x+sin⁡ x)=lim⁡x→π4(tan⁡ x)+lim⁡x→π4(sin⁡ x)=tan⁡ π4+sin⁡ π4=1+22=2+22  (Answer)\begin{align*} \lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x+\sin\:x\right) & =\lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x\right)+\lim\limits_{x\to \frac{\pi }{4}}\left(\sin\:x\right)\\ & =\tan\:\frac{\pi }{4}+\sin\:\frac{\pi }{4}\\ & =1+\frac{\sqrt{2}}{2}\\ & =\frac{2+\sqrt{2}}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 2

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PROBLEM:

Evaluate lim⁡x→3(4x+2x+4)\displaystyle \lim\limits_{x\to 3}\left(\frac{4x+2}{x+4}\right).

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SOLUTION:

lim⁡x→3(4x+2x+4)=lim⁡x→3(4x+2)lim⁡x→3(x+4)=lim⁡x→3(4x)+lim⁡x→3(2)lim⁡x→3(x)+lim⁡x→3(4)=4⋅lim⁡x→3(x)+23+4=4⋅3+23+4=12+27=147=2  (Answer)\begin{align*} \lim_{x\to 3}\left(\frac{4x+2}{x+4}\right)& =\frac{\lim\limits_{x\to 3}\left(4x+2\right)}{\lim\limits_{x\to 3}\left(x+4\right)}\\ & =\frac{\lim\limits_{x\to 3}\left(4x\right)+\lim\limits_{x\to 3}\left(2\right)}{\lim\limits_{x\to 3}\left(x\right)+\lim\limits_{x\to 3}\left(4\right)}\\ & =\frac{4\cdot \lim\limits_{x\to 3}\left(x\right)+2}{3+4}\\ & =\frac{4\cdot 3+2}{3+4}\\ & =\frac{12+2}{7}\\ & =\frac{14}{7}\\ & =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 1

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PROBLEM:

Evaluate lim⁡x→2(x2−4x+3)\displaystyle \lim _{x\to 2}\left(x^2-4x+3\right).

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SOLUTION:

lim⁡x→2(x2−4x+3)=lim⁡x→2(x2)−lim⁡x→2(4x)+lim⁡x→2(3)=[lim⁡x→2(x)]2−4lim⁡x→2(x)+3=(2)2−4(2)+3=−1  (Answer)\begin{align*} \lim_{x\to 2}\left(x^2-4x+3\right)& = \lim_{x\to 2}\left(x^2\right)-\lim_{x\to 2}\left(4x\right)+\lim_{x\to 2}\left(3\right)\\ & =\left[\lim_{x\to 2}\left(x\right)\right]^2-4\lim_{x\to 2}\left(x\right)+3\\ & =\left(2\right)^2-4\left(2\right)+3\\ & =-1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 10

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PROBLEM:

If  f(x)=4x+3\displaystyle f\left(x\right)=\frac{4}{x+3} and  g(x)=x2−3\displaystyle \:g\left(x\right)=x^2-3 , find f[g(x)]\displaystyle f\left[g\left(x\right)\right] and g[f(x)]\displaystyle g\left[f\left(x\right)\right].

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SOLUTION:

Part A

f[g(x)]=4(x2−3)+3=4x2  (Answer)\begin{align*} f\left[g\left(x\right)\right] & =\frac{4}{\left(x^2-3\right)+3}\\ & =\frac{4}{x^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part B

g[f(x)]=(4x+3)2−3=16(x+3)2−3=16−3(x+3)2(x+3)2=16−3(x2+6x+9)(x+3)2=16−3x2−18x−27(x+3)2=−3x2−18x−11(x+3)2  (Answer)\begin{align*} g\left[f\left(x\right)\right] & =\left(\frac{4}{x+3}\right)^2-3\\ & =\frac{16}{\left(x+3\right)^2}-3\\ & =\frac{16-3\left(x+3\right)^2}{\left(x+3\right)^2}\\ & =\frac{16-3\left(x^2+6x+9\right)}{\left(x+3\right)^2}\\ & =\frac{16-3x^2-18x-27}{\left(x+3\right)^2}\\ & =\frac{-3x^2-18x-11}{\left(x+3\right)^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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