Tag Archives: math solution

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 5

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PROBLEM:

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

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SOLUTION:

Plug in the value x=8.

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

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SOLUTION:

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

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

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SOLUTION:

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

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SOLUTION:

\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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