Tag Archives: math solution

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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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 2

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PROBLEM:

Evaluate \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=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^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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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 1

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PROBLEM:

Evaluate \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 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{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 \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. 

\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 \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.

\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 \displaystyle \lim_{x\to 2}\left(4x-3\right)\left(x^2+5\right).

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SOLUTION:

Plug the value x=2.

\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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