Difference between revisions of "009A Sample Final A, Problem 6"

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|More work is required here.  Since we need to find the limits at <math style="vertical-align: 0%;">\pm\infty</math>, we can multiply our <math style="vertical-align: -20%;">f(x)</math> by
 
|More work is required here.  Since we need to find the limits at <math style="vertical-align: 0%;">\pm\infty</math>, we can multiply our <math style="vertical-align: -20%;">f(x)</math> by
 
|-
 
|-
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math>
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math>
 
|-
 
|-
|This expression is equal to 1 for positive values of ''x'', and is equal to -1 for negative values of ''x''.  Since multiplying ''f''(''x'') by an expression equal to 1 doesn't change the limit, we will add a negative sign to it when considering the limit as x goes to <math style="vertical-align: -5%;">-\infty</math>.  Thus,
+
|<br>This expression is equal to 1 for positive values of ''x'', and is equal to -1 for negative values of ''x''.  Since multiplying ''f''(''x'') by an expression equal to 1 doesn't change the limit, we will add a negative sign to it when considering the limit as x goes to <math style="vertical-align: -5%;">-\infty</math>.  Thus,
 
|-
 
|-
 
|<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}</math>
 
|<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}</math>

Revision as of 21:40, 26 March 2015


6. Find the vertical and horizontal asymptotes of the function  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\frac{\sqrt{4x^{2}+3}}{10x-20}.}

Foundations:  
Vertical asymptotes occur whenever the denominator of a rational function goes to zero, and  it doesn't cancel from the numerator.
On the other hand, horizontal asymptotes represent the limit as x goes to either positive or negative infinity.

 Solution:

Vertical Asymptotes:  
Setting the denominator to zero, we have
     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 = 10x-20 = 10(x-2),}
which has a root at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = 2.} This is our vertical asymptote.
Horizontal Asymptotes:  
More work is required here. Since we need to find the limits at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm\infty} , we can multiply our Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} by

     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.}

This expression is equal to 1 for positive values of x, and is equal to -1 for negative values of x. Since multiplying f(x) by an expression equal to 1 doesn't change the limit, we will add a negative sign to it when considering the limit as x goes to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\infty} . Thus,

     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}}

Thus, we have a horizontal asymptote at y = -1/5 on the left (as x goes to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\infty} ), and a horizontal asymptote at y = 1/5 as x goes to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty} ).

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