standard 9 homework\nscore: 4/27 answered: 4/20\nprogress sav\nquestion 5\nfind the horizontal asymptote of…
standard 9 homework\nscore: 4/27 answered: 4/20\nprogress sav\nquestion 5\nfind the horizontal asymptote of ( f(x)=\frac{3 x+x^{3}-4}{5 x^{3}+4 x^{2}-5} ). if the horizontal asymptote does not exist, enter dne.\nthe horizontal asymptote is ( y=) \nquestion help: video message instructor post to forum\nsubmit question
Answer
Explanation:
Step1: Divide numerator and denominator by (x^3)
$$\lim_{x\rightarrow\pm\infty}\frac{3x + x^3 - 4}{5x^3 + 4x^2 - 5}=\lim_{x\rightarrow\pm\infty}\frac{\frac{3x}{x^3}+\frac{x^3}{x^3}-\frac{4}{x^3}}{\frac{5x^3}{x^3}+\frac{4x^2}{x^3}-\frac{5}{x^3}}$$
Step2: Simplify the expression
$$=\lim_{x\rightarrow\pm\infty}\frac{\frac{3}{x^2}+1-\frac{4}{x^3}}{5+\frac{4}{x}-\frac{5}{x^3}}$$
Step3: Evaluate the limit
As (x\rightarrow\pm\infty), (\frac{3}{x^2}\rightarrow0), (\frac{4}{x^3}\rightarrow0), (\frac{4}{x}\rightarrow0), (\frac{5}{x^3}\rightarrow0) $$=\frac{0 + 1-0}{5+0 - 0}=\frac{1}{5}$$
Answer:
(\frac{1}{5})