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 | ||
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| − | | <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math> | + | |<br> <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math> |
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| − | |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, |
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|<br> <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> <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
| 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 |
| which has a root at This is our vertical asymptote. |
| Horizontal Asymptotes: |
|---|
| More work is required here. Since we need to find the limits at , we can multiply our by |
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 . Thus, |
Thus, we have a horizontal asymptote at y = -1/5 on the left (as x goes to ), and a horizontal asymptote at y = 1/5 as x goes to ). |
