﻿ Limits at Infinity---Roots and Absolute Values - 8 Practice Problems explained step by step with interactive problems, showing all work.

Limits at Infinity---Roots and Absolute Values: Practice Problems

Practice Problems

Step 1

Simplify the absolute value.

Since the limit examines positive $$x$$-values, we know $$|x| = x$$.

$$\displaystyle\lim_{x\to\infty}\,\frac{4\blue{|x|} + 3}{9x -2} % = \lim_{x\to\infty}\,\frac{4\blue x + 3}{9x -2}$$

Step 2

Factor the $$x$$ out of the numerator and denominator. Then divide out the common factor.

\begin{align*}% \lim_{x\to\infty}\,\frac{4\blue x + 3}{9\blue x -2} % & = \lim_{x\to\infty}\,\frac{% \blue x\left(% 4 + \frac 3 x% \right)% } {% \blue x\left(% 9 - \frac 2 x% \right)% } \\[6pt] % & = \lim_{x\to\infty}\,\frac{4 + \frac 3 x}{9 - \frac 2 x} \end{align*}

Step 3

Evaluate the limit.

$$\displaystyle\lim_{x\to\infty}\,\frac{4 + \blue{\frac 3 x}}{9 - \red{\frac 2 x}} % = \frac{4 + \blue 0}{9 - \red 0} % = \frac 4 9$$

$$\displaystyle \lim_{x\to\infty}\,\frac{4|x| + 3}{9x -2} = \frac 4 9$$

Step 1

Simplify the absolute values.

Since the limit examines negative $$x$$-values, we know $$|x| = -x$$.

\begin{align*}% \lim_{x\to-\infty}\,\frac{% 2\blue{|x|}-15% } {% 6-\red{|x|}% } % & = \lim_{x\to-\infty}\,\frac{2\blue{(-x)}-15}{6-\red{(-x)}} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{\blue{-2x}-15}{6\red{+x}} \end{align*}

Step 2

Factor an $$x$$ out of the numerator and denominator. Then divide out the common factor.

\begin{align*}% \lim_{x\to-\infty}\,\frac{-\blue x-15}{6+\blue x}% % & = \lim_{x\to-\infty}\,\frac{% \blue x\left(% -2-\frac{15} x% \right) } {% \blue x\left(% \frac 6 x+1% \right) } \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% -2-\frac{15} x% } {% \frac 6 x+1% } \end{align*}

Step 3

Evaluate the limit.

$$\displaystyle\lim_{x\to-\infty}\,\frac{% -2-\blue{\frac{15} x}% } {% \red{\frac 6 x}+1% } % = \frac{-2-\blue 0}{\red 0+1} % = -2$$

$$\displaystyle \lim_{x\to-\infty}\,\frac{2|x|-15}{6-|x|} = - 2$$

Step 1

Factor the $$x^2$$ out of the square-root.

\begin{align*} \lim_{x\to\infty}\,\frac{3x+\sqrt{\blue{x^2}+1}}{2x+3} % & = \lim_{x\to\infty}\,\frac{% 3x+\sqrt{% \blue{x^2}\left(1+ \frac 1 {x^2}\right) } }{2x+3}\\[6pt] % & = \lim_{x\to\infty}\,\frac{% 3x+\sqrt{\blue{x^2}} \cdot \sqrt{1+ \frac 1 {x^2}} } {2x+3} \\[6pt] % & = \lim_{x\to\infty}\,\frac{% 3x+\blue{|x|} \cdot \sqrt{1+ \frac 1 {x^2}} } {2x+3} \end{align*}

Step 2

Simplify the absolute value.

Since the limit examines positive value of $$x$$, we know $$|x| = x$$.

$$\displaystyle\lim_{x\to\infty}\,\frac{% 3x+\blue{|x|}% \cdot \sqrt{1+ \frac 1 {x^2}}% } {2x+3} % = \lim_{x\to\infty}\,\frac{% 3x+\blue x% \cdot \sqrt{1+ \frac 1 {x^2}}% } {2x+3}$$

Step 3

Factor an $$x$$ out of the numerator and the denominator. Then divide out the common factor.

\begin{align*}% \lim_{x\to\infty}\,\frac{% 3\blue x+\blue x% \cdot \sqrt{1+ \frac 1 {x^2}} } {2\blue x+3} % & = \lim_{x\to\infty}\,\frac{% \blue x \left(% 3+\sqrt{1+ \frac 1 {x^2}}% \right) } {% \blue x\left(% 2+\frac 3 x% \right) } \\[6pt] % & = \lim_{x\to\infty}\,\frac{% 3+\sqrt{1+ \frac 1 {x^2}} } {2+\frac 3 x} \end{align*}

Step 4

Evaluate the limit.

$$\displaystyle\lim_{x\to\infty}\,\frac{% 3+\sqrt{% 1+ \blue{\frac 1 {x^2}} } } {2+\red{\frac 3 x}} % = \frac{% 3+\sqrt{1+ \blue 0}% } {2+\red 0} % = \frac 4 2 % = 2$$

$$\displaystyle \lim_{x\to\infty}\,\frac{3x+\sqrt{x^2+1}}{2x+3} = 2$$

Step 1

Factor the $$x^2$$ out of the square-root.

\begin{align*} & \quad \lim_{x\to-\infty}\,\frac{8x+\sqrt{16\blue{x^2}+7}}{3 - 5x}\\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 8x+\sqrt{% \blue{x^2}% \left(% 16+\frac 7 {x^2}% \right) } } {3 - 5x} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 8x+\sqrt{% \blue{x^2}% }% \cdot \sqrt{% 16+\frac 7 {x^2}% } } {3 - 5x}% \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 8x+\blue{|x|}% \cdot \sqrt{% 16+\frac 7 {x^2}% } } {3 - 5x} \end{align*}

Step 2

Simplify the absolute value.

Since the limit examines negative $$x$$-values, we know $$|x| = -x$$.

\begin{align*}% & \quad \lim_{x\to-\infty}\,\frac{% 8x+\blue{|x|}% \cdot \sqrt{16+\frac 7 {x^2}}% } {3 - 5x}\\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 8x+\blue{(-x)}% \cdot \sqrt{16+\frac 7 {x^2}}% } {3 - 5x} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 8x \blue{-x}% \cdot \sqrt{16+\frac 7 {x^2}}% } {3 - 5x} \end{align*}

Step 3

Factor an $$x$$ out of the numerator and denominator. Then divide out the common factor.

\begin{align*}% & \quad \lim_{x\to-\infty}\, \frac{% 8\blue x - \blue x\cdot% \sqrt{% 16+\frac 7 {x^2} }% } {3 - 5\blue x}\\[6pt] % & = \lim_{x\to-\infty}\, \frac{% \blue x% \left(% 8 - % \sqrt{% 16+\frac 7 {x^2} }% \right) }{% \blue x% \left(% \frac 3 x - 5% \right) } \\[6pt] % & = \lim_{x\to-\infty}\, \frac{% 8 - % \sqrt{% 16+\frac 7 {x^2} } } {\frac 3 x - 5} \end{align*}

Step 4

Evaluate the limit.

\begin{align*}% \lim_{x\to-\infty}\, \frac{% 8 - % \sqrt{% 16+\blue{\frac 7 {x^2}} } } {% \red{\frac 3 x} - 5 } % & = \frac{% 8 - \sqrt{16+\blue 0} } {\red 0 - 5} \\[6pt] % & = \frac{ 8 -4}{-5} \\[6pt] % & = -\frac 4 5 \end{align*}

$$\displaystyle \lim_{x\to-\infty}\,\frac{8x+\sqrt{16x^2+7}}{3 - 5x} = -\frac 4 5$$

Step 1

Factor the $$x^2$$ out of the square-root.

\begin{align*} \lim_{x\to\infty}\, \frac{% 7x-% \sqrt{% 5\blue{x^2}+17% } } {10x-7} % & = \lim_{x\to\infty}\, \frac{% 7x-% \sqrt{% \blue{x^2}% \left(% 5+\frac{17}{x^2} \right) } } {10x-7} \\[6pt] % & = \lim_{x\to\infty}\,% \frac{% 7x-\sqrt{\blue{x^2}}\cdot% \sqrt{% 5+\frac{17}{x^2}% } } {10x-7} \\[6pt] % & = \lim_{x\to\infty}\,% \frac{% 7x-\blue{|x|}\cdot% \sqrt{% 5+\frac{17}{x^2}% } } {10x-7} \end{align*}

Step 2

Simplify the absolute value.

Since the limit examines positive values of $$x$$, we know $$|x| = x$$.

$$\displaystyle\lim_{x\to\infty}\,% \frac{% 7x-\blue{|x|}\cdot \sqrt{% 5+\frac{17}{x^2} } } {10x-7} % = \lim_{x\to\infty}\,% \frac{% 7x-\blue x\cdot% \sqrt{% 5+\frac{17}{x^2}% } } {10x-7}$$

Step 3

Factor an $$x$$ out of the numerator and denominator. Divide out the common factor.

\begin{align*}% \lim_{x\to\infty}\,\frac{% 7\blue x-\blue x\cdot% \sqrt{5+\frac{17}{x^2}% } } {10\blue x-7} % & = \lim_{x\to\infty}\,\frac{% \blue x\left(% 7-\sqrt{5+\frac{17}{x^2}}% \right)% } {\blue x\left(10 - \frac 7 {x^2}\right)}% \\[6pt] % & = \lim_{x\to\infty}\,\frac{% 7-\sqrt{5+\frac{17}{x^2}}% } {10 - \frac 7 {x^2}} \end{align*}

Step 4

Evaluate the limit.

$$\displaystyle\lim_{x\to\infty}\,\frac{% 7-\sqrt{% 5+\blue{\frac{17}{x^2}} } }{10 - \red{\frac 7 {x^2}}} % = \frac{7-\sqrt{5+\blue 0}}{10 - \red 0} % = \frac{7-\sqrt 5}{10}$$

$$\displaystyle \lim_{x\to\infty}\,\frac{7x-\sqrt{5x^2+17}}{10x-7} = \frac{7 - \sqrt 5}{10}$$

Step 1

Factor the $$x^2$$ out of the square-root.

\begin{align*} & \quad \lim_{x\to-\infty}\,\frac{2x-\sqrt{\blue{x^2}+5x+6}}{12x+3} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 2x-\sqrt{% \blue{x^2}\left(1+\frac 5 x+\frac 6 {x^2}\right) } }{12x+3}\\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 2x-\sqrt{\blue{x^2}}\cdot \sqrt{1+\frac 5 x+\frac 6 {x^2}} }{12x+3}\\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 2x-\blue{|x|}\cdot \sqrt{1+\frac 5 x+\frac 6 {x^2}} }{12x+3} \end{align*}

Step 2

Simplify the absolute value.

Since the limit examines negative values of $$x$$, we know $$|x| = -x$$.

\begin{align*}% & \quad \lim_{x\to-\infty}\,\frac{% 2x-\blue{|x|}\cdot \sqrt{1+\frac 5 x+\frac 6 {x^2}} } {12x+3}\\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 2x-\blue{(-x)}\cdot \sqrt{1+\frac 5 x+\frac 6 {x^2}} } {12x+3} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 2x \blue{+x}\cdot \sqrt{1+\frac 5 x+\frac 6 {x^2}} } {12x+3} \end{align*}

Step 3

Factor an $$x$$ out of the numerator and denominator. Then divide out the common factor.

\begin{align*}% & \quad \lim_{x\to-\infty}\,\frac{% 2x+x\cdot \sqrt{1+\frac 5 x+\frac 6 {x^2}} } {12x+3} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% \blue x \left(% 2+\sqrt{1+\frac 5 x+\frac 6 {x^2}} \right) } {\blue x\left(12+\frac 3 x\right)} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 2+\sqrt{1+\frac 5 x+\frac 6 {x^2}} }{12+\frac 3 x} \end{align*}.

Step 4

Evaluate the limit.

\begin{align*}% \lim_{x\to-\infty}\,\frac{% 2+\sqrt{% 1+\blue{\frac 5 x}+\blue{\frac 6 {x^2}} } }{12+\red{\frac 3 x}} % & = \frac{2+\sqrt{1+\blue 0+\blue 0}}{12+\red 0} \\[6pt] % & = \frac 3 {12} \\[6pt] % & = \frac 1 4 \end{align*}

$$\displaystyle \lim_{x\to-\infty}\,\frac{2x-\sqrt{x^2+5x+6}}{12x+3} = \frac 1 4$$

Step 1

In the numerator and denominator, factor the $$x^2$$ out of the square-roots.

\begin{align*} \lim_{x\to\infty}\,\frac{3x+\sqrt{4\blue{x^2}+1}}{9x-\sqrt{\red{x^2}+3}} % & = \lim_{x\to\infty}\,\frac{% 3x+\sqrt{\blue{x^2}\left(4+\frac 1 {x^2}\right)} }{% 9x-\sqrt{\red{x^2}\left(1+\frac 3 {x^2}\right)} }\\[6pt] % & = \lim_{x\to\infty}\,\frac{% 3x+\sqrt{\blue{x^2}}\cdot \sqrt{4+\frac 1 {x^2}} }{% 9x-\sqrt{\red{x^2}}\cdot \sqrt{1+\frac 3 {x^2}} }\\[6pt] % & = \lim_{x\to\infty}\,\frac{% 3x+\blue{|x|}\cdot \sqrt{4+\frac 1 {x^2}} }{% 9x-\red{|x|}\cdot \sqrt{1+\frac 3 {x^2}} } \end{align*}

Step 2

Simplify the absolute values.

Since the limit examines positive $$x$$-values, we know $$|x| = x$$.

$$\displaystyle\lim_{x\to\infty}\,\frac{% 3x+\blue{|x|}\cdot \sqrt{4+\frac 1 {x^2}} }{% 9x-\red{|x|}\cdot \sqrt{1+\frac 3 {x^2}} } % = \lim_{x\to\infty}\,\frac{% 3x+\blue x\cdot \sqrt{4+\frac 1 {x^2}} }{% 9x-\red x\cdot \sqrt{1+\frac 3 {x^2}} }$$

Step 3

Factor an $$x$$ out of the numerator and denominator. Then divide out the common factor.

\begin{align*}% \lim_{x\to\infty}\,\frac{% 3x+x\cdot \sqrt{4+\frac 1 {x^2}} }{% 9x-x\cdot \sqrt{1+\frac 3 {x^2}} } % & = \lim_{x\to\infty}\,\frac{% \blue x\left(3 + \sqrt{4+\frac 1 {x^2}}\right) }{% \blue x\left(9-\sqrt{1+\frac 3 {x^2}}\right) } \\[6pt] % & = \lim_{x\to\infty}\,\frac{% 3 + \sqrt{4+\frac 1 {x^2}} }{% 9-\sqrt{1+\frac 3 {x^2}} } \end{align*}

Step 4

Evaluate the limit.

$$\displaystyle\lim_{x\to\infty}\,\frac{% 3 + \sqrt{% 4+\blue{\frac 1 {x^2}} } }{% 9-\sqrt{% 1+\red{\frac 3 {x^2}} } } % = \frac{3 + \sqrt{4+\blue 0} } {9-\sqrt{1+\red 0}} % = \frac 5 8$$

$$\displaystyle \lim_{x\to\infty}\,\frac{3x+\sqrt{4x^2+1}}{9x-\sqrt{x^2+3}} = \frac 5 8$$

Step 1

Factor the $$x^2$$ out of the square-root.

\begin{align*} & \quad \lim_{x\to-\infty}\,\frac{6x+\sqrt{25\blue{x^2}+8}}{3|x|+7} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 6x+\sqrt{% \blue{x^2}\left(25+\frac 8 {x^2}\right) } }{3|x|+7}\\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 6x+\sqrt{\blue{x^2}}\cdot \sqrt{25+\frac 8 {x^2}} }{3|x|+7}\\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 6x+\blue{|x|}\cdot \sqrt{25+\frac 8 {x^2}} }{3|x|+7} \end{align*}

Step 2

Simplify the absolute values.

Since the limit examines negative $$x$$-values, we know $$|x| = -x$$.

\begin{align*}% & \quad \lim_{x\to-\infty}\,\frac{% 6x+\blue{|x|}\cdot \sqrt{25+\frac 8 {x^2}} }{3\red{|x|}+7} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 6x+\blue{(-x)}\cdot \sqrt{25+\frac 8 {x^2}} }{3\red{(-x)}+7} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 6x-x\cdot \sqrt{25+\frac 8 {x^2}} }{-3x+7} \end{align*}

Step 3

Factor an $$x$$ out of the numerator and denominator. Then divide out the common factor.

\begin{align*}% & \quad \lim_{x\to-\infty}\,\frac{% 6x-x\cdot \sqrt{25+\frac 8 {x^2}} }{-3x+7} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% \blue x \left(6-\sqrt{25+\frac 8 {x^2}}\right) }{\blue x\left(-3+\frac 7 x\right)} \\[6pt] % & = \lim_{x\to-\infty}\,\frac{% 6-\sqrt{25+\frac 8 {x^2}} }{-3+\frac 7 x} \end{align*}

Step 4

Evaluate the limit.

\begin{align*}% \lim_{x\to-\infty}\,\frac{% 6-\sqrt{25+\blue{\frac 8 {x^2}}} }{-3+\red{\frac 7 x}} % & = \frac{6-\sqrt{25+\blue 0}}{-3+\red 0} \\[6pt] % & = \frac{6 -5}{-3} \\[6pt] % & = -\frac 1 3 \end{align*}

$$\displaystyle \lim_{x\to-\infty}\,\frac{6x+\sqrt{25x^2+8}}{3|x|+7} = -\frac 1 3$$