37 - 50. horizontal asymptotes determine $lim_{x\rightarrowinfty}f(x)$ and $lim_{x\rightarrow-infty}f(x)$…

37 - 50. horizontal asymptotes determine $lim_{x\rightarrowinfty}f(x)$ and $lim_{x\rightarrow-infty}f(x)$ for the following functions. then give the horizontal asymptotes of $f$ (if any).\n37. $f(x)=\frac{4x}{20x + 1}$\n38. $f(x)=\frac{3x^{2}-7}{x^{2}+5x}$\n39. $f(x)=\frac{6x^{2}-9x + 8}{3x^{2}+2}$\n40. $f(x)=\frac{12x^{8}-3}{3x^{8}-2x^{7}}$\n41. $f(x)=\frac{3x^{3}-7}{x^{4}+5x^{2}}$\n42. $f(x)=\frac{2x + 1}{3x^{4}-2}$\n43. $f(x)=\frac{40x^{5}+x^{2}}{16x^{4}-2x}$\n44. $f(x)=\frac{6x^{2}+1}{sqrt{4x^{4}+3x + 1}}$

Answer

For (f(x)=\frac{4x}{20x + 1}):
- Find (\lim_{x\rightarrow\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x)
    - (\lim_{x\rightarrow\infty}\frac{4x}{20x + 1}=\lim_{x\rightarrow\infty}\frac{4}{20+\frac{1}{x}})
  - Step 2: Evaluate the limit as (x\rightarrow\infty)
    - As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{1}{x}=0). So (\lim_{x\rightarrow\infty}\frac{4}{20+\frac{1}{x}}=\frac{4}{20}=\frac{1}{5})
- Find (\lim_{x\rightarrow-\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x) (note that for (x\rightarrow-\infty), (\frac{4x}{x}=4) and (\frac{20x + 1}{x}=20+\frac{1}{x}))
    - (\lim_{x\rightarrow-\infty}\frac{4x}{20x + 1}=\lim_{x\rightarrow-\infty}\frac{4}{20+\frac{1}{x}})
  - Step 2: Evaluate the limit as (x\rightarrow-\infty)
    - As (x\rightarrow-\infty), (\lim_{x\rightarrow-\infty}\frac{1}{x}=0). So (\lim_{x\rightarrow-\infty}\frac{4}{20+\frac{1}{x}}=\frac{4}{20}=\frac{1}{5})
- Horizontal - asymptote: The horizontal asymptote is (y = \frac{1}{5})
For (f(x)=\frac{3x^{2}-7}{x^{2}+5x}):
- Find (\lim_{x\rightarrow\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{2})
    - (\lim_{x\rightarrow\infty}\frac{3x^{2}-7}{x^{2}+5x}=\lim_{x\rightarrow\infty}\frac{3-\frac{7}{x^{2}}}{1 + \frac{5}{x}})
  - Step 2: Evaluate the limit as (x\rightarrow\infty)
    - As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{7}{x^{2}} = 0) and (\lim_{x\rightarrow\infty}\frac{5}{x}=0). So (\lim_{x\rightarrow\infty}\frac{3-\frac{7}{x^{2}}}{1+\frac{5}{x}}=3)
- Find (\lim_{x\rightarrow-\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{2})
    - (\lim_{x\rightarrow-\infty}\frac{3x^{2}-7}{x^{2}+5x}=\lim_{x\rightarrow-\infty}\frac{3-\frac{7}{x^{2}}}{1+\frac{5}{x}})
  - Step 2: Evaluate the limit as (x\rightarrow-\infty)
    - As (x\rightarrow-\infty), (\lim_{x\rightarrow-\infty}\frac{7}{x^{2}} = 0) and (\lim_{x\rightarrow-\infty}\frac{5}{x}=0). So (\lim_{x\rightarrow-\infty}\frac{3-\frac{7}{x^{2}}}{1+\frac{5}{x}}=3)
- Horizontal - asymptote: The horizontal asymptote is (y = 3)
For (f(x)=\frac{6x^{2}-9x + 8}{3x^{2}+2}):
- Find (\lim_{x\rightarrow\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{2})
    - (\lim_{x\rightarrow\infty}\frac{6x^{2}-9x + 8}{3x^{2}+2}=\lim_{x\rightarrow\infty}\frac{6-\frac{9}{x}+\frac{8}{x^{2}}}{3+\frac{2}{x^{2}}})
  - Step 2: Evaluate the limit as (x\rightarrow\infty)
    - As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{9}{x}=0), (\lim_{x\rightarrow\infty}\frac{8}{x^{2}} = 0) and (\lim_{x\rightarrow\infty}\frac{2}{x^{2}}=0). So (\lim_{x\rightarrow\infty}\frac{6-\frac{9}{x}+\frac{8}{x^{2}}}{3+\frac{2}{x^{2}}}=2)
- Find (\lim_{x\rightarrow-\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{2})
    - (\lim_{x\rightarrow-\infty}\frac{6x^{2}-9x + 8}{3x^{2}+2}=\lim_{x\rightarrow-\infty}\frac{6-\frac{9}{x}+\frac{8}{x^{2}}}{3+\frac{2}{x^{2}}})
  - Step 2: Evaluate the limit as (x\rightarrow-\infty)
    - As (x\rightarrow-\infty), (\lim_{x\rightarrow-\infty}\frac{9}{x}=0), (\lim_{x\rightarrow-\infty}\frac{8}{x^{2}} = 0) and (\lim_{x\rightarrow-\infty}\frac{2}{x^{2}}=0). So (\lim_{x\rightarrow-\infty}\frac{6-\frac{9}{x}+\frac{8}{x^{2}}}{3+\frac{2}{x^{2}}}=2)
- Horizontal - asymptote: The horizontal asymptote is (y = 2)
For (f(x)=\frac{12x^{8}-3}{3x^{8}-2x^{7}}):
- Find (\lim_{x\rightarrow\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{8})
    - (\lim_{x\rightarrow\infty}\frac{12x^{8}-3}{3x^{8}-2x^{7}}=\lim_{x\rightarrow\infty}\frac{12-\frac{3}{x^{8}}}{3-\frac{2}{x}})
  - Step 2: Evaluate the limit as (x\rightarrow\infty)
    - As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{3}{x^{8}} = 0) and (\lim_{x\rightarrow\infty}\frac{2}{x}=0). So (\lim_{x\rightarrow\infty}\frac{12-\frac{3}{x^{8}}}{3-\frac{2}{x}} = 4)
- Find (\lim_{x\rightarrow-\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{8})
    - (\lim_{x\rightarrow-\infty}\frac{12x^{8}-3}{3x^{8}-2x^{7}}=\lim_{x\rightarrow-\infty}\frac{12-\frac{3}{x^{8}}}{3-\frac{2}{x}})
  - Step 2: Evaluate the limit as (x\rightarrow-\infty)
    - As (x\rightarrow-\infty), (\lim_{x\rightarrow-\infty}\frac{3}{x^{8}} = 0) and (\lim_{x\rightarrow-\infty}\frac{2}{x}=0). So (\lim_{x\rightarrow-\infty}\frac{12-\frac{3}{x^{8}}}{3-\frac{2}{x}} = 4)
- Horizontal - asymptote: The horizontal asymptote is (y = 4)
For (f(x)=\frac{3x^{3}-7}{x^{4}+5x^{2}}):
- Find (\lim_{x\rightarrow\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{4})
    - (\lim_{x\rightarrow\infty}\frac{3x^{3}-7}{x^{4}+5x^{2}}=\lim_{x\rightarrow\infty}\frac{\frac{3}{x}-\frac{7}{x^{4}}}{1+\frac{5}{x^{2}}})
  - Step 2: Evaluate the limit as (x\rightarrow\infty)
    - As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{3}{x}=0), (\lim_{x\rightarrow\infty}\frac{7}{x^{4}} = 0) and (\lim_{x\rightarrow\infty}\frac{5}{x^{2}}=0). So (\lim_{x\rightarrow\infty}\frac{\frac{3}{x}-\frac{7}{x^{4}}}{1+\frac{5}{x^{2}}}=0)
- Find (\lim_{x\rightarrow-\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{4})
    - (\lim_{x\rightarrow-\infty}\frac{3x^{3}-7}{x^{4}+5x^{2}}=\lim_{x\rightarrow-\infty}\frac{\frac{3}{x}-\frac{7}{x^{4}}}{1+\frac{5}{x^{2}}})
  - Step 2: Evaluate the limit as (x\rightarrow-\infty)
    - As (x\rightarrow-\infty), (\lim_{x\rightarrow-\infty}\frac{3}{x}=0), (\lim_{x\rightarrow-\infty}\frac{7}{x^{4}} = 0) and (\lim_{x\rightarrow-\infty}\frac{5}{x^{2}}=0). So (\lim_{x\rightarrow-\infty}\frac{\frac{3}{x}-\frac{7}{x^{4}}}{1+\frac{5}{x^{2}}}=0)
- Horizontal - asymptote: The horizontal asymptote is (y = 0)
For (f(x)=\frac{2x + 1}{3x^{4}-2}):
- Find (\lim_{x\rightarrow\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{4})
    - (\lim_{x\rightarrow\infty}\frac{2x + 1}{3x^{4}-2}=\lim_{x\rightarrow\infty}\frac{\frac{2}{x^{3}}+\frac{1}{x^{4}}}{3-\frac{2}{x^{4}}})
  - Step 2: Evaluate the limit as (x\rightarrow\infty)
    - As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{2}{x^{3}} = 0), (\lim_{x\rightarrow\infty}\frac{1}{x^{4}} = 0) and (\lim_{x\rightarrow\infty}\frac{2}{x^{4}}=0). So (\lim_{x\rightarrow\infty}\frac{\frac{2}{x^{3}}+\frac{1}{x^{4}}}{3-\frac{2}{x^{4}}}=0)
- Find (\lim_{x\rightarrow-\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{4})
    - (\lim_{x\rightarrow-\infty}\frac{2x + 1}{3x^{4}-2}=\lim_{x\rightarrow-\infty}\frac{\frac{2}{x^{3}}+\frac{1}{x^{4}}}{3-\frac{2}{x^{4}}})
  - Step 2: Evaluate the limit as (x\rightarrow-\infty)
    - As (x\rightarrow-\infty), (\lim_{x\rightarrow-\infty}\frac{2}{x^{3}} = 0), (\lim_{x\rightarrow-\infty}\frac{1}{x^{4}} = 0) and (\lim_{x\rightarrow-\infty}\frac{2}{x^{4}}=0). So (\lim_{x\rightarrow-\infty}\frac{\frac{2}{x^{3}}+\frac{1}{x^{4}}}{3-\frac{2}{x^{4}}}=0)
- Horizontal - asymptote: The horizontal asymptote is (y = 0)
For (f(x)=\frac{40x^{5}+x^{2}}{16x^{4}-2x}):
- Find (\lim_{x\rightarrow\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{4})
    - (\lim_{x\rightarrow\infty}\frac{40x^{5}+x^{2}}{16x^{4}-2x}=\lim_{x\rightarrow\infty}\frac{40x+\frac{1}{x^{2}}}{16-\frac{2}{x^{3}}})
  - Step 2: Evaluate the limit as (x\rightarrow\infty)
    - As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0) and (\lim_{x\rightarrow\infty}\frac{2}{x^{3}}=0). And (\lim_{x\rightarrow\infty}(40x+\frac{1}{x^{2}})=\infty), so (\lim_{x\rightarrow\infty}\frac{40x+\frac{1}{x^{2}}}{16-\frac{2}{x^{3}}}=\infty)
- Find (\lim_{x\rightarrow-\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{4})
    - (\lim_{x\rightarrow-\infty}\frac{40x^{5}+x^{2}}{16x^{4}-2x}=\lim_{x\rightarrow-\infty}\frac{40x+\frac{1}{x^{2}}}{16-\frac{2}{x^{3}}})
  - Step 2: Evaluate the limit as (x\rightarrow-\infty)
    - As (x\rightarrow-\infty), (\lim_{x\rightarrow-\infty}\frac{1}{x^{2}} = 0) and (\lim_{x\rightarrow-\infty}\frac{2}{x^{3}}=0). And (\lim_{x\rightarrow-\infty}(40x+\frac{1}{x^{2}})=-\infty), so (\lim_{x\rightarrow-\infty}\frac{40x+\frac{1}{x^{2}}}{16-\frac{2}{x^{3}}}=-\infty). There is no horizontal asymptote.
For (f(x)=\frac{6x^{2}+1}{\sqrt{4x^{4}+3x + 1}}):
- Find (\lim_{x\rightarrow\infty}f(x)):
  - Step 1: Divide numerator and denominator by (x^{2}) (note that (\sqrt{4x^{4}+3x + 1}=x^{2}\sqrt{4+\frac{3}{x^{3}}+\frac{1}{x^{4}}}) for (x>0))
    - (\lim_{x\rightarrow\infty}\frac{6x^{2}+1}{\sqrt{4x^{4}+3x + 1}}=\lim_{x\rightarrow\infty}\frac{6+\frac{1}{x^{2}}}{\sqrt{4+\frac{3}{x^{3}}+\frac{1}{x^{4}}}})
  - Step 2: Evaluate the limit as (x\rightarrow\infty)
    - As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0), (\lim_{x\rightarrow\infty}\frac{3}{x^{3}} = 0) and (\lim_{x\rightarrow\infty}\frac{1}{x^{4}}=0). So (\lim_{x\rightarrow\infty}\frac{6+\frac{1}{x^{2}}}{\sqrt{4+\frac{3}{x^{3}}+\frac{1}{x^{4}}}}=\frac{6}{\sqrt{4}} = 3)
- Find (\lim_{x\rightarrow-\infty}f(x)):
  - **Step 1: Divide numerator and denominator by (x^{2}) (note that (\sqrt{4x^{4}+3x + 1}=-x^{2}\sqrt{4+\frac{3}{x^{3}}+\frac{1}{x^{4