Question

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

Answer 1

$$\begin{gathered} \underset{x\to -1/2}{\lim }\left[\frac{8x^3+1}{2x+1}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to -1/2}{\lim }\left[\frac{{\left(2x\right)}^3+1}{2x+1}\right] \\ =\underset{x\to -1/2}{\lim }\left[\frac{\left(2x+1\right)\left\{{\left(2x\right)}^2-2x\times 1+1^2\right\}}{\left(2x+1\right)}\right]\left[\because A^3+B^3=\left(A+B\right)\left(A^2-AB+B^2\right)\right] \\ =\underset{x\to -1/2}{\lim }\left[{\left(2x\right)}^2-2x+1\right] \\ ={\left(2\times \frac{-1}{2}\right)}^2-2\times \frac{-1}{2}+1 \\ =1+1+1 \\ =3 \end{gathered}$$