Key to "Divide numerator and denominator by the highest power..."

alesterp

Answered question

2021-10-17

Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits

lim_{x \to - \infty} {(\frac{x^{2} + x - 1}{8 x^{2} - 3})}^{\frac{1}{3}}

escumantsu

Skilled2021-10-18Added 98 answers

Given,

lim_{x \to - \infty} {(\frac{x^{2} + x - 1}{8 x^{2} - 3})}^{\frac{1}{3}}

Concept used:

{(a^{m} b)}^{n} = a^{m n} {(b)}^{n}

Apply the above concept, we get

lim_{x \to - \infty} {(\frac{x^{2} + x - 1}{8 x^{2} - 3})}^{\frac{1}{3}} = lim_{x \to - \infty} (\frac{x^{\frac{2}{3}}}{x^{\frac{2}{3}}}) {(\frac{1 + \frac{1}{x} - \frac{1}{x^{2}}}{8 - \frac{3}{x^{2}}})}^{\frac{1}{3}}

= lim_{x \to - \infty} (1) {(\frac{1 + \frac{1}{x} - \frac{1}{x^{2}}}{8 - \frac{3}{x^{2}}})}^{\frac{1}{3}}

= {(\frac{1 + \frac{1}{- \infty} - \frac{1}{{(- \infty)}^{2}}}{8 - \frac{3}{{(- \infty)}^{2}}})}^{\frac{1}{3}}

= {(\frac{1 - 0 - 0}{8 - 0})}^{\frac{1}{3}}

= {(\frac{1}{8})}^{\frac{1}{3}}

= {(\frac{1}{2^{3}})}^{\frac{1}{3}}

= \frac{1}{2}

Jeffrey Jordon

Expert2022-06-29Added 2605 answers

Answer is given below (on video)

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