divide numerator and denominator by the highest power of x in the denominator and proceed from there. find…
divide numerator and denominator by the highest power of x in the denominator and proceed from there. find $lim_{x\rightarrowinfty}\frac{4sqrt{x}+x^{-7}}{3x - 2}$. write $infty$ or $-infty$ where appropriate.\n$lim_{x\rightarrowinfty}\frac{4sqrt{x}+x^{-7}}{3x - 2}=square$ (simplify your answer.)
Answer
Explanation:
Step1: Identify highest - power of x
The highest power of x in the denominator (3x - 2) is (x^1).
Step2: Divide numerator and denominator by x
[ \begin{align*} \lim_{x\rightarrow\infty}\frac{4\sqrt{x}+x^{-7}}{3x - 2}&=\lim_{x\rightarrow\infty}\frac{\frac{4\sqrt{x}}{x}+\frac{x^{-7}}{x}}{\frac{3x}{x}-\frac{2}{x}}\ &=\lim_{x\rightarrow\infty}\frac{4x^{-\frac{1}{2}}+x^{-8}}{3 - 2x^{-1}} \end{align*} ]
Step3: Use limit rules
We know that (\lim_{x\rightarrow\infty}x^{-n}=0) for (n>0). So (\lim_{x\rightarrow\infty}4x^{-\frac{1}{2}} = 0), (\lim_{x\rightarrow\infty}x^{-8}=0) and (\lim_{x\rightarrow\infty}2x^{-1}=0). [ \begin{align*} \lim_{x\rightarrow\infty}\frac{4x^{-\frac{1}{2}}+x^{-8}}{3 - 2x^{-1}}&=\frac{\lim_{x\rightarrow\infty}(4x^{-\frac{1}{2}}+x^{-8})}{\lim_{x\rightarrow\infty}(3 - 2x^{-1})}\ &=\frac{0 + 0}{3-0} \end{align*} ]
Answer:
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