 # Find each of the following limits. If the limit is not finite, indicate or for one- or two-sided limits, as appropriate. lim_{xrightarrowinfty}frac{4x^3-2x-1}{x^2-1} Elleanor Mckenzie 2021-03-09 Answered
Find each of the following limits. If the limit is not finite, indicate or for one- or two-sided limits, as appropriate.
$\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2x-1}{{x}^{2}-1}$
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We need to compute the value of given limit
$\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2x-1}{{x}^{2}-1}$
Dividing both numerator and denominator by ${x}^{2}$ , we get
$\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{4{x}^{3}-2x-1}{{x}^{2}}}{\frac{{x}^{2}-1}{{x}^{2}}}$
$=\underset{x\to \mathrm{\infty }}{lim}\frac{4x-\frac{2}{x}-\frac{1}{{x}^{2}}}{1-\frac{1}{{x}^{2}}}$
$=\frac{\underset{x\to \mathrm{\infty }}{lim}\left(4x-\frac{2}{x}-\frac{1}{{x}^{2}}\right)}{\underset{x\to \mathrm{\infty }}{lim}\left(1-\frac{1}{{x}^{2}}\right)}$
$=\frac{\left(\mathrm{\infty }-0-0\right)}{\left(1-0\right)}$
$=\mathrm{\infty }$
Answer: $\therefore \underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2x-1}{{x}^{2}-1}=\mathrm{\infty }$
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