Lim x -1-28 x 3-1-2 x -1

Lim_x → 1/28x^3+1/2x+1 It is of the form 0/0 =lim_x → 1/28(x^3+1/8)/2(x+1/2) =8/2lim_x → 1/2(x^3+(1/2)^3)/x+1/2 =4 lim_x → 1/2(x+1/2)(x^2+1/2 1/4x)/(x+1/2) ...

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Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

lim x →-1/28 ...

Question

${lim}_{x \to - \frac{1}{2}} \frac{8 x^{3} + 1}{2 x + 1}$

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Solution

${lim}_{x \to - \frac{1}{2}} \frac{8 x^{3} + 1}{2 x + 1}$

It is of the form $\frac{0}{0}$

$= {lim}_{x \to - \frac{1}{2}} \frac{8 (x^{3} + \frac{1}{8})}{2 (x + \frac{1}{2})}$

$= \frac{8}{2} {lim}_{x \to - \frac{1}{2}} \frac{(x^{3} + (\frac{1}{2})^{3})}{x + \frac{1}{2}}$

$= 4 {lim}_{x \to - \frac{1}{2}} \frac{(x + \frac{1}{2}) (x^{2} + \frac{1}{2} - \frac{1}{4} x)}{(x + \frac{1}{2})}$

$= 4 {lim}_{x \to - \frac{1}{2}} (x^{2} - \frac{1}{2} x + \frac{1}{4})$

$= 4 ({(\frac{- 1}{2})}^{2} + \frac{1}{4} - \frac{1}{2} (\frac{- 1}{2}))$

$= 4 (\frac{1}{4} + \frac{1}{4} + \frac{2}{4})$

=3

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limx→−128x3+12x+1

limx→−128x3+12x+1 It is of the form 00 =limx→−128(x3+18)2(x+12) =82limx→−12(x3+(12)3)x+12 =4limx→−12(x+12)(x2+12−14x)(x+12) =4limx→−12(x2−12x+14) =4((−12)2+14−12(−12)) =4(14+14+24) =3

${lim}_{x \to - \frac{1}{2}} \frac{8 x^{3} + 1}{2 x + 1}$