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entonces
![\sqrt[3]{2x+3}^3-1^3=\left(\sqrt[3]{2x+3}-1\right)\left(\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1\right)](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%2B3%7D%5E3-1%5E3%3D%5Cleft%28%5Csqrt%5B3%5D%7B2x%2B3%7D-1%5Cright%29%5Cleft%28%5Csqrt%5B3%5D%7B2x%2B3%7D%5E2%2B%5Csqrt%5B3%5D%7B2x%2B3%7D%2B1%5Cright%29 "\sqrt[3]{2x+3}^3-1^3=\left(\sqrt[3]{2x+3}-1\right)\left(\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1\right)")
Luego
![\displaystyle
L=\lim\limits_{x\to-1} \frac{x+1}{\sqrt[3]{2x+3}-1} \\ \\
L=\lim\limits_{x\to-1} \frac{x+1}{\sqrt[3]{2x+3}-1}\cdot \frac{\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1}{\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1} \\ \\
L=\lim\limits_{x\to-1} \frac{(x+1)(\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1)}{2x+3-1} \\ \\
L=\lim\limits_{x\to-1} \frac{(x+1)(\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1)}{2(x+1)}\\ \\
L=\lim\limits_{x\to-1} \frac{\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1}{2} \\ \\](https://tex.z-dn.net/?f=%5Cdisplaystyle%0AL%3D%5Clim%5Climits_%7Bx%5Cto-1%7D+%5Cfrac%7Bx%2B1%7D%7B%5Csqrt%5B3%5D%7B2x%2B3%7D-1%7D+%5C%5C+%5C%5C%0AL%3D%5Clim%5Climits_%7Bx%5Cto-1%7D+%5Cfrac%7Bx%2B1%7D%7B%5Csqrt%5B3%5D%7B2x%2B3%7D-1%7D%5Ccdot+%5Cfrac%7B%5Csqrt%5B3%5D%7B2x%2B3%7D%5E2%2B%5Csqrt%5B3%5D%7B2x%2B3%7D%2B1%7D%7B%5Csqrt%5B3%5D%7B2x%2B3%7D%5E2%2B%5Csqrt%5B3%5D%7B2x%2B3%7D%2B1%7D+%5C%5C+%5C%5C%0AL%3D%5Clim%5Climits_%7Bx%5Cto-1%7D+%5Cfrac%7B%28x%2B1%29%28%5Csqrt%5B3%5D%7B2x%2B3%7D%5E2%2B%5Csqrt%5B3%5D%7B2x%2B3%7D%2B1%29%7D%7B2x%2B3-1%7D+%5C%5C+%5C%5C%0AL%3D%5Clim%5Climits_%7Bx%5Cto-1%7D+%5Cfrac%7B%28x%2B1%29%28%5Csqrt%5B3%5D%7B2x%2B3%7D%5E2%2B%5Csqrt%5B3%5D%7B2x%2B3%7D%2B1%29%7D%7B2%28x%2B1%29%7D%5C%5C+%5C%5C%0AL%3D%5Clim%5Climits_%7Bx%5Cto-1%7D+%5Cfrac%7B%5Csqrt%5B3%5D%7B2x%2B3%7D%5E2%2B%5Csqrt%5B3%5D%7B2x%2B3%7D%2B1%7D%7B2%7D+%5C%5C+%5C%5C "\displaystyle
L=\lim\limits_{x\to-1} \frac{x+1}{\sqrt[3]{2x+3}-1} \\ \\
L=\lim\limits_{x\to-1} \frac{x+1}{\sqrt[3]{2x+3}-1}\cdot \frac{\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1}{\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1} \\ \\
L=\lim\limits_{x\to-1} \frac{(x+1)(\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1)}{2x+3-1} \\ \\
L=\lim\limits_{x\to-1} \frac{(x+1)(\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1)}{2(x+1)}\\ \\
L=\lim\limits_{x\to-1} \frac{\sqrt[3]{2x+3}^2+\sqrt[3]{2x+3}+1}{2} \\ \\")
![\displaystyle
L=\frac{\sqrt[3]{2(-1)+3}^2+\sqrt[3]{2(-1)+3}+1}{2} \\ \\
\boxed{L=\frac{3}{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%0AL%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B2%28-1%29%2B3%7D%5E2%2B%5Csqrt%5B3%5D%7B2%28-1%29%2B3%7D%2B1%7D%7B2%7D+%5C%5C+%5C%5C%0A%5Cboxed%7BL%3D%5Cfrac%7B3%7D%7B2%7D%7D "\displaystyle
L=\frac{\sqrt[3]{2(-1)+3}^2+\sqrt[3]{2(-1)+3}+1}{2} \\ \\
\boxed{L=\frac{3}{2}}")
entonces
Luego
Dostoievski1:
nesse final como é que ³√[2(-1)+3²] dá 1?
pra soma mais o 1 que dá na outra raiz e 1 solto para dá 3
en realidad es: ³√[2(-1)+3]² = 1
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