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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Tag Archives: math solution

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 7

PROBLEM:

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

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 6

PROBLEM:

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

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 5

PROBLEM:

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

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 4

PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right)$ .

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 3

PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\right)$ .

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

PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to 2}\left(\frac{x^2+2x-8}{3x-6}\right)$

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 1

PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)$

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 8

PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right)$ .

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 7

PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{3x}}{x\sqrt{x+1}}\right)$ .

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 6

PROBLEM:

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

PROBLEM:

Evaluate lim⁡x→1 x−1x+3−2 \displaystyle \lim\limits_{x\to 1}\:\frac{x-1}{\sqrt{x+3}-2}x→1lim​x+3​−2x−1​.

PROBLEM:

Evaluate lim⁡x→0 x+16−4x \displaystyle\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}x→0lim​xx+16​−4​.

PROBLEM:

Evaluate lim⁡x→0 (x+3)2−92x\displaystyle \lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x}x→0lim​2x(x+3)2−9​.

PROBLEM:

Evaluate lim⁡x→2(x3−x2−x−22x3−5x2+5x−6) \displaystyle \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right)x→2lim​(2x3−5x2+5x−6x3−x2−x−2​).

PROBLEM:

Evaluate lim⁡x→3(x3−13x+12x3−14x+15) \displaystyle \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\right)x→3lim​(x3−14x+15x3−13x+12​).

PROBLEM:

Evaluate lim⁡x→2(x2+2x−83x−6)\displaystyle \lim\limits_{x\to 2}\left(\frac{x^2+2x-8}{3x-6}\right)x→2lim​(3x−6x2+2x−8​)

PROBLEM:

Evaluate lim⁡x→4(x3−64x2−16)\displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)x→4lim​(x2−16x3−64​)

PROBLEM:

Evaluate lim⁡x→0(3x+2x2−2x+4)\displaystyle \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right)x→0lim​(x2−2x+43x+2​).

PROBLEM:

Evaluate lim⁡x→3(3xxx+1)\displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{3x}}{x\sqrt{x+1}}\right)x→3lim​(xx+1​3x​​).

PROBLEM:

Evaluate lim⁡x→2(4x−3)(x2+5)\displaystyle \lim_{x\to 2}\left(4x-3\right)\left(x^2+5\right)x→2lim​(4x−3)(x2+5).

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

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

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

Evaluate $\displaystyle \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right)$ .

Evaluate $\displaystyle \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\right)$ .

Evaluate $\displaystyle \lim\limits_{x\to 2}\left(\frac{x^2+2x-8}{3x-6}\right)$

Evaluate $\displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)$

Evaluate $\displaystyle \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right)$ .

Evaluate $\displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{3x}}{x\sqrt{x+1}}\right)$ .

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