Evaluate the following limit:$$\underset{x\to -\mathrm{\infty}}{lim}\frac{8{x}^{4}+3{x}^{2}-x}{\sqrt{9{x}^{8}-2{x}^{5}+4}}$$

Aubrie Mccall
2022-10-01
Answered

Evaluate the following limit:$$\underset{x\to -\mathrm{\infty}}{lim}\frac{8{x}^{4}+3{x}^{2}-x}{\sqrt{9{x}^{8}-2{x}^{5}+4}}$$

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xgirlrogueim

Answered 2022-10-02
Author has **13** answers

Solution:

$$\underset{x\to -\mathrm{\infty}}{lim}\frac{8{x}^{4}+3{x}^{2}-x}{\sqrt{9{x}^{8}-2{x}^{5}+4}}\phantom{\rule{0ex}{0ex}}=\underset{x\to -\mathrm{\infty}}{lim}\frac{{x}^{4}(8+\frac{3}{{x}^{2}}-\frac{1}{{x}^{3}})}{{x}^{4}\sqrt{(9-\frac{2}{{x}^{3}}+\frac{4}{{x}^{8}})}}\phantom{\rule{0ex}{0ex}}=\underset{x\to -\mathrm{\infty}}{lim}\frac{(8+\frac{3}{{x}^{2}}-\frac{1}{{x}^{2}})}{\sqrt{9-\frac{2}{{x}^{3}}+\frac{4}{{x}^{8}}}}\phantom{\rule{0ex}{0ex}}=\frac{8+0-0}{\sqrt{9-0+0}}\phantom{\rule{0ex}{0ex}}=\frac{8}{\sqrt{9}}=\frac{8}{3}$$

$$\underset{x\to -\mathrm{\infty}}{lim}\frac{8{x}^{4}+3{x}^{2}-x}{\sqrt{9{x}^{8}-2{x}^{5}+4}}\phantom{\rule{0ex}{0ex}}=\underset{x\to -\mathrm{\infty}}{lim}\frac{{x}^{4}(8+\frac{3}{{x}^{2}}-\frac{1}{{x}^{3}})}{{x}^{4}\sqrt{(9-\frac{2}{{x}^{3}}+\frac{4}{{x}^{8}})}}\phantom{\rule{0ex}{0ex}}=\underset{x\to -\mathrm{\infty}}{lim}\frac{(8+\frac{3}{{x}^{2}}-\frac{1}{{x}^{2}})}{\sqrt{9-\frac{2}{{x}^{3}}+\frac{4}{{x}^{8}}}}\phantom{\rule{0ex}{0ex}}=\frac{8+0-0}{\sqrt{9-0+0}}\phantom{\rule{0ex}{0ex}}=\frac{8}{\sqrt{9}}=\frac{8}{3}$$

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