Tag Archives: Feliciano and Uy

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 12

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right)

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SOLUTION: 

A straight substitution of x=0 leads to the indeterminate form 0\cdot 0 which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

\begin{align*}
\lim\limits_{x\to \:0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right) & =\lim\limits_{x\to 0}\left(\frac{1\cdot \:\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)}{x}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{\frac{\sqrt{x+9}-3}{3\sqrt{x+9}}}{x}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right)\cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} \\ \\
& =\lim\limits_{x\to 0}\left(\frac{x}{3x\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{1}{3\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\
& =\frac{1}{3\left(\sqrt{0+9}+3\right)\sqrt{0+9}} \\ \\
& =\frac{1}{54} \ \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 11

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Evaluate \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=3 leads to the indeterminate form \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows.

\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 \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=2 leads to the indeterminate form \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

\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 \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=4 leads to the indeterminate form \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows:

\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 \displaystyle \lim\limits_{x\to 8}\:\frac{\sqrt[3]{x}-2}{x-8}.

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SOLUTION:

A straight substitution of x=8 leads to the indeterminate form \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

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

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 SOLUTION:

A straight substitution of  x=1 leads to the indeterminate form \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

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

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 SOLUTION:

A straight substitution of x=0 leads to the indeterminate form \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows.

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

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 SOLUTION:

A straight substitution of x=0 leads to the indeterminate form \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

\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  \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=2 leads to the indeterminate form \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

\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  \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=3 leads to the indeterminate form \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows.

\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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