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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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