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lim-x-1-7-x-3-1-3-3-x-2-x-1-




Question Number 138997 by bramlexs22 last updated on 21/Apr/21
lim_(x→1) ((((7+x^3 ))^(1/3) −(√(3+x^2 )))/(x−1))=?
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{7}+\mathrm{x}^{\mathrm{3}} }−\sqrt{\mathrm{3}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}−\mathrm{1}}=? \\ $$
Answered by EDWIN88 last updated on 21/Apr/21
 lim_(x→1)  ((x^2 (7+x^3 )^(−(2/3)) −x(3+x^2 )^(−(1/2)) )/1)  = ((8^(−(2/3)) −4^(−(1/2)) )/1)= (1/4)−(1/2)=−(1/4)
$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{7}+\mathrm{x}^{\mathrm{3}} \right)^{−\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{x}\left(\mathrm{3}+\mathrm{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}} \\ $$$$=\:\frac{\mathrm{8}^{−\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}}=\:\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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