Difference between revisions of "009A Sample Final A, Problem 6"
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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> | ||
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− | |<br>Thus, we have a horizontal asymptote at <math style="vertical-align: - | + | |<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -24%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -1%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -24%;">y=1/5</math> on the right (as <math style="vertical-align: -2%;">x</math> goes to <math style="vertical-align: 0%;">+\infty</math>). |
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[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']] |