Lim x- 1 - 2 8x3 - 12x - 1

lim_x→1/28x^3+1/2x+1=8(1/2)^3+1/2(1/2)+1 =8(1/8)+1/2=2/2=1 ...

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L'Hospital Rule to Remove Indeterminate Form

lim x→ 1 / 2 ...

Question

$l i m x \to \frac{1}{2}$ $\frac{8 x^{3} + 1}{2 x + 1}$

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Solution

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(2 x + \frac{1}{x})

8 x^{3} + 12 x + \frac{6}{x} + \frac{1}{x^{3}}

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

t a n^{- 1} [\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}] - t a n^{- 1} [\frac{4 x}{1 - 4 x^{2}}] = t a n^{- 1} 2 x, | 2 x | < \frac{1}{\sqrt{3}}

x + \frac{1}{4 x} = \frac{3}{2}

8 x^{3} + \frac{1}{8 x^{3}}

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