Author Archives: Engineering Math

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 21

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PROBLEM:

If \displaystyle f\left(x\right)=x^2-2x+3, find \displaystyle \lim\limits_{x\to 2}\left(\frac{f\left(x\right)-f\left(2\right)}{x-2}\right).

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SOLUTION:

\begin{align*}
\displaystyle \lim\limits_{x\to 2}\left(\displaystyle \frac{f\left(x\right)-f\left(2\right)}{x-2}\right) & =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-\left(2^2-2\cdot 2+3\right)}{x-2}\right) \\
\\
& =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-3}{x-2}\right)\\
\\
& =\lim\limits_{x\to 2}\left(\displaystyle \frac{x^2-2x}{x-2}\right)\\
\\

\end{align*}

Direct substitution of x=2 gives the indeterminate form \frac{0}{0}. Therefore, we proceed by factoring the numerator.

\begin{align*}
& =\lim\limits_{x\to 2}\left(\displaystyle \frac{x\left(x-2\right)}{x-2}\right) \\
\\
& =\lim\limits_{x\to 2}\left(x\right) \\
\\
& =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 20

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PROBLEM:

If \displaystyle f\left(x\right)=\sqrt{x}, find \displaystyle \lim\limits_{x\to 0}\left(\frac{f\left(9+x\right)-f\left(9\right)}{x}\right).

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SOLUTION:

\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(9+x\right)-f\left(9\right)}{x}\right)=\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-\sqrt{9}}{x}\right)

Direct substitution of x=0 gives the indeterminate form \frac{0}{0}. Therefore, we proceed by rationalizing the numerator.

\begin{align*}
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-3}{x}\cdot \displaystyle \frac{\sqrt{9+x}+3}{\sqrt{9+x}+3}\right)\\
\\
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{9+x-9}{x\left(\sqrt{9+x}+3\right)}\right)\\
\\
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{x}{x\left(\sqrt{9+x}+3\right)}\right)\\
\\
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{1}{\left(\sqrt{9+x}+3\right)}\right)\\
\\
& =\left(\displaystyle \frac{1}{\left(\sqrt{9+0}+3\right)}\right)\\
\\
& =\displaystyle \frac{1}{6} \ \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 19

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PROBLEM:

If f\left(x\right)=\sqrt{x}, find \displaystyle \lim\limits_{x\to 4}\left(\frac{f\left(x\right)-f\left(4\right)}{x-4}\right)

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SOLUTION:

\displaystyle \lim\limits_{x\to 4}\left(\displaystyle \frac{f\left(x\right)-f\left(4\right)}{x-4}\right)=\lim\limits_{x\to 4}\left(\displaystyle \frac{\sqrt{x}-\sqrt{4}}{x-4}\right)

Direct substitution of x=4 gives the indeterminate form \frac{0}{0}. Therefore, we proceed by rationalizing the numerator.

\begin{align*}
& =\lim\limits_{x\to 4}\left(\displaystyle \frac{\sqrt{x}-2}{x-4}\right)\cdot \displaystyle \frac{\sqrt{x}+2}{\sqrt{x}+2} \\
\\
& =\lim\limits_{x\to 4}\left(\displaystyle \frac{x-4}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right) \\
\\
& =\lim\limits_{x\to 4}\left(\displaystyle \frac{1}{\sqrt{x}+2}\right) \\
\\
& =\left(\displaystyle \frac{1}{\sqrt{4}+2}\right) \\
\\
& =\displaystyle \frac{1}{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 18

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to \pi }\left(\frac{\sin^2\left(x\right)}{1+\cos\left(x\right)}\right).

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SOLUTION:

Direct substitution of x=\pi gives the indeterminate form \frac{0}{0}. Therefore, we should apply trigonometric identities.

We know the Pythagorean identity, \sin^2\left(x\right)=1-\cos^2\left(x\right). Therefore, we have

\begin{align*}
\displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{\sin^2\left(x\right)}{1+\cos\left(x\right)}\right) & = \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{1-\cos^2\left(x\right)}{1+\cos\left(x\right)}\right) \\
\\
& = \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{\left(1+\cos\left(x\right)\right)\left(1-\cos\left(x\right)\right)}{1+\cos\left(x\right)}\right)\\
\\
& =\lim\limits_{x\to \pi }\left(1-\cos\left(x\right)\right)\\
\\
& =\left(1-\cos\left(\pi \right)\right)\\
\\
& =\left(1-\left(-1\right)\right)\\
\\
& =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 17

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right).

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SOLUTION:

Direct substitution of x=0 gives the indeterminate form \frac{0}{0}. Therefore, we should apply trigonometric identities.

We know that \sin\left(2x\right)=2\sin\left(x\right)\cos\left(x\right), so we can rewrite the original function as

\begin{align*}
\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =\lim\limits_{x\to 0}\left(\frac{\sin\left(x\right)\cdot 2\left(\sin\left(x\right) \cos\left(x\right)\right)}{1-\cos\left(x\right)}\right)\\
\\
& =\displaystyle  2\cdot \lim\limits_{x\to 0}\left(\frac{\sin^2\left(x\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\
\end{align*}

We also know the Pythagorean identity \sin^2\left(x\right)=1-\cos^2\left(x\right). So,

\begin{align*}
\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1-\cos^2\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\
\\
& =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1+\cos\left(x\right)\right)\left(1-\cos\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right) \\
\\
& =2\cdot \lim\limits_{x\to 0}\left(\left(1+\cos\left(x\right)\right)\cos\left(x\right)\right) \\
\\
& = 2\cdot \left(\left(1+\cos\left(0\right)\right)\cos\left(0\right)\right) \\
\\
& =2\cdot \left(\left(1+1\right)\cdot 1\right) \\
\\
& =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 16

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 0}\left(\frac{1-\cos^2\left(x\right)}{1+\cos\left(x\right)}\right)

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SOLUTION:

This problem can be solved using a direct substitution of x=0. That is

\begin{align*}
\lim\limits_{x\to 0}\left(\frac{1-\cos^2  x }{1+\cos x }\right) & =\frac{1-\cos^2\left(0\right)}{1+\cos\left(0\right)} \\
\\
& =\frac{1-1}{1+1}\\ 
\\
& = 0 \ \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 15

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PROBLEM:

Evaluate \displaystyle \lim_{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right).

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SOLUTION:

A straight substitution of \displaystyle x=\frac{\pi }{4} leads to the indeterminate form \displaystyle \frac{0}{0} which is meaningless.

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

\begin{align*}
\displaystyle \lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right) & =\lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\frac{\sin x}{\cos x}}\right) \\ 
\\
& =\lim _{x\to 0}\left(\frac{\sin^3 x}{\frac{\sin x \cos x - \sin x}{\cos x}}\right) \\
\\
& = \lim _{x\to 0}\left(\frac{\sin^3 x \cos x }{\sin x \cos x - \sin x}\right) \\
\\
&=\lim _{x\to \:0}\left(\frac{\sin^3 x \cos x}{\left(\sin x \right)\left(\cos x-1\right)}\right) \\
\\
& =\lim _{x\to 0}\left(\frac{\sin^2 x \cos x }{\left(\cos x -1\right)}\right) \\
\\
& =\lim _{x\to 0}\left(\frac{\left(1-\cos^2 x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\
\\
& =\lim\limits_{x\to 0}\left(\frac{\left(1+\cos x \right) \left(1-\cos x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\
\\
& =\lim _{x\to 0}\left(\frac{\left(1+\cos x \right) \cdot \cos\left(x\right)}{-1}\right) \\
\\
& =-1\cdot \lim _{x\to 0}\left(\left(1+\cos x\right)\cdot\cos x \right) \\
\\
& =-1\cdot \left(1+\cos 0 \right)\cdot \cos 0 \\
\\
& =-1\cdot \left(1+1\right)\cdot 1 \\
\\
& =-2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 14

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to \frac{\pi }{4}}\left(\frac{\tan\:2x}{\sec\:2x}\right).

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SOLUTION:

A straight substitution of x=\frac{\pi }{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*}
\displaystyle \lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\frac{\tan\:2x}{\sec\:2x}\right) & =\lim\limits_{x\to \frac{\pi }{4}}\left(\frac{\frac{\sin\:2x}{\cos\:2x}}{\frac{1}{\cos\:2x}}\right) \\
\\
& =\lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\sin\:2x\right) \\
\\
& =\sin\left(2\cdot \frac{\pi }{4}\right) \\
\\
& =\sin\frac{\pi }{2} \\
\\
& =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 13

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\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{\sqrt{x^2-9}}{x-3}\right) & =\lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\cdot \frac{\sqrt{x^2-9}}{\sqrt{x^2-9}}\right) \\
\\
& =\lim\limits_{x\to 3}\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\sqrt{x^2-9}}\right) \\
\\
& =\lim\limits_{x\to 3}\left(\frac{x^2-9}{\left(x-3\right)\sqrt{x^2-9}}\right) \\
\\
& =\lim _{x\to 3}\left(\frac{x+3}{\sqrt{x^2-9}}\right) \\
\\
& =\frac{3+3}{\sqrt{3^2-9}} \\
\\
& =\frac{6}{0} \\
\\
& =\infty \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Since the function’s limit is different from the left to its limits from the right, the limit does not exist. 

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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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