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Question Number 198123 by a.lgnaoui last updated on 10/Oct/23

$$\mathrm{Determiner}$$$${\lim }_{\mathrm{x}\to 3}\frac{\mathbf{x}-3}{{}^3\sqrt{\mathbf{x}+5}-2}$$$$$$

Answered by Mathspace last updated on 10/Oct/23

$$a^3-b^3=(a-b)(a^2+ab+b^2)\Rightarrow$$$$a-b=\left({}^3\sqrt{a}{-}^3\sqrt{b}\right)({\left({}^3\sqrt{a}\right)}^2{+}^3\sqrt{ab}+{\left({}^3\sqrt{b}\right)}^2)\Rightarrow$$$$\left({}^3\sqrt{x+5}\right)-\left({}^3\sqrt{8}\right)$$$$=\frac{x+5-8}{{(x+5)}^{\frac{2}{3}}+{\left(8(x+5)\right)}^{\frac{1}{3}}+8^{\frac{2}{3}}}$$$$\Rightarrow {lim}_{x\to 3}f\left(x\right)$$$$={lim}_{x\to 3}\frac{x-3}{\frac{x-3}{{(x+5)}^{\frac{2}{3}}+2{(x+5)}^{\frac{1}{3}}+4}}$$$$={lim}_{x\to 3}{(x+5)}^{\frac{2}{3}}+2{(x+5)}^{\frac{1}{3}}+4$$$$=8^{\frac{2}{3}}+2\times 8^{\frac{1}{3}}+4$$$$=4+4+4=12$$

Commented by a.lgnaoui last updated on 10/Oct/23

$$\mathrm{exact}$$

Answered by MM42 last updated on 10/Oct/23

$$\sqrt[3]{x+5}-2=u\Rightarrow x+5={(u+2)}^3\Rightarrow x={(u+2)}^3-5$$$$\Rightarrow {lim}_{u\to 0}\frac{{(u+2)}^3-8}{u}={lim}_{u\to 0}\frac{u(u^2+6u+12)}{u}=12✓$$

Answered by cortano12 last updated on 11/Oct/23

$$\underset{―}{\underbrace{}}\underbrace{\mathcal{Y}}3$$

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