Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 119216 by benjo_mathlover last updated on 23/Oct/20

$$\underset{x\to 1}{\lim }\frac{x^3-3x+2}{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}?$$

Commented by MJS_new last updated on 23/Oct/20

$$\mathrm{we}\mathrm{can}‘‘{\mathrm{hospitalize}}^{″}\mathrm{it}2\mathrm{times},\mathrm{answer}\mathrm{is}27$$

Commented by benjo_mathlover last updated on 23/Oct/20

$$yes...$$

Commented by bemath last updated on 23/Oct/20

$$without{L}^{\prime }Hopital$$$$\underset{x\to 1}{\lim }\frac{(x-1)(x^2+x-2)}{{(\sqrt[3]{x}-1)}^2}=\underset{x\to 1}{\lim }\frac{{(x-1)}^2(x+2)}{{(\sqrt[3]{x}-1)}^2}$$$$=3\times \underset{x\to 1}{\lim }{\left[\frac{x-1}{\sqrt[3]{x}-1}\right]}^2$$$$=3\times {\left[\underset{x\to 1}{\lim }\frac{(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x}-1}\right]}^2$$$$=3\times {\left(3\right)}^2=27$$

Commented by malwan last updated on 23/Oct/20

$$cooool$$

Answered by benjo_mathlover last updated on 23/Oct/20

$$let\sqrt[3]{x}=t$$$$\underset{t\to 1}{\lim }\frac{t^9-3t^3+2}{t^2-2t+1}=\underset{t\to 1}{\lim }\frac{9t^8-9t^2}{2t-2}$$$$\underset{t\to 1}{\lim }\frac{72t^7-18t}{2}=36-9=27$$

Answered by 1549442205PVT last updated on 23/Oct/20

$$\mathrm{Put}{}^3\sqrt{\mathrm{x}}=\mathrm{u}(\mathrm{x}\to 1\sim \mathrm{u}\to 1)$$$$\underset{\mathrm{x}\to 1}{\lim }\frac{x^3-3x+2}{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}=\underset{\mathrm{u}\to 1}{\lim }\frac{{\mathrm{u}}^9-{3\mathrm{u}}^3+2}{{\mathrm{u}}^2-2\mathrm{u}+1}$$$$\iff \underset{\mathrm{u}\to 1}{\lim }\frac{({\mathrm{u}}^7+{2\mathrm{u}}^6+{3\mathrm{u}}^5+{4\mathrm{u}}^4+{5\mathrm{u}}^3+{6\mathrm{u}}^2+4\mathrm{u}+2){(\mathrm{u}-1)}^2}{{(\mathrm{u}-1)}^2}$$$$=\underset{\mathrm{u}\to 1}{\lim }({\mathrm{u}}^7+{2\mathrm{u}}^6+{3\mathrm{u}}^5+{4\mathrm{u}}^4+{5\mathrm{u}}^3+{6\mathrm{u}}^2+4\mathrm{u}+2)$$$$=1+2+3+4+5+6+4+2=27$$

Answered by Dwaipayan Shikari last updated on 23/Oct/20

$$\underset{x\to 1}{\lim }\frac{(x+2){(x-1)}^2}{{(x^{\frac{1}{3}}-1)}^2}=(x+2){(x^{\frac{2}{3}}+x^{\frac{1}{3}}+1)}^2=3.9=27$$

Answered by mathmax by abdo last updated on 24/Oct/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^3-3\mathrm{x}+2}{{\left({}^3\sqrt{\mathrm{x}}\right)}^2-2\left({}^3\sqrt{\mathrm{x}}\right)+1}\mathrm{we}\mathrm{have}$$$${\mathrm{x}}^3-3\mathrm{x}+2={\mathrm{x}}^3-\mathrm{x}-2\mathrm{x}+2=\mathrm{x}({\mathrm{x}}^2-1)-2(\mathrm{x}-1)$$$$=\mathrm{x}(\mathrm{x}-1)(\mathrm{x}+1)-2(\mathrm{x}-1)=(\mathrm{x}-1)({\mathrm{x}}^2+\mathrm{x}-2)\mathrm{and}$$$${\left({}^3\sqrt{\mathrm{x}}\right)}^2-2\left({}^3\sqrt{\mathrm{x}}\right)+1={({}^3\sqrt{\mathrm{x}}-1)}^2\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{(\mathrm{x}-1)({\mathrm{x}}^2+\mathrm{x}-2)}{{({}^3\sqrt{\mathrm{x}}-1)}^2}$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}{}^3\sqrt{\mathrm{x}}-1=\mathrm{t}{\Rightarrow }^3\sqrt{\mathrm{x}}=\mathrm{t}+1\Rightarrow \mathrm{x}={(\mathrm{t}+1)}^3\Rightarrow$$$$(\mathrm{x}\to 1\Rightarrow \mathrm{t}\to 0)\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left({(\mathrm{t}+1)}^3\right)=$$$$=\frac{({(\mathrm{t}+1)}^3-1)({(\mathrm{t}+1)}^6+{(\mathrm{t}+1)}^3-2)}{{\mathrm{t}}^2}$$$$=\frac{({\mathrm{t}}^3+{3\mathrm{t}}^2+3\mathrm{t})({(\mathrm{t}+1)}^6+{(\mathrm{t}+1)}^2-2)}{{\mathrm{t}}^2}$$$$=\frac{({\mathrm{t}}^2+3\mathrm{t}+3)({(\mathrm{t}+1)}^6+{(\mathrm{t}+1)}^3-2)}{\mathrm{t}}\mathrm{we}\mathrm{have}$$$${(\mathrm{t}+1)}^6+{(\mathrm{t}+1)}^3-2=\sum _{\mathrm{k}=0}^6{\mathrm{C}}_6^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}+{\mathrm{t}}^3+{3\mathrm{t}}^2+3\mathrm{t}-1$$$$=1+{\mathrm{C}}_6^1\mathrm{t}+{\mathrm{C}}_6^2{\mathrm{t}}^2+{\mathrm{C}}_6^3{\mathrm{t}}^3+{\mathrm{C}}_6^4{\mathrm{t}}^4+{\mathrm{C}}_6^5{\mathrm{t}}^5+{\mathrm{t}}^6+{\mathrm{t}}^3+{3\mathrm{t}}^2+3\mathrm{t}-1$$$$=9\mathrm{t}+(3+{\mathrm{C}}_6^2){\mathrm{t}}^2+(1+{\mathrm{C}}_6^3){\mathrm{t}}^3+{\mathrm{C}}_6^4{\mathrm{t}}^4+{\mathrm{C}}_6^5{\mathrm{t}}^5+{\mathrm{t}}^6\Rightarrow$$$$\mathrm{f}\left({(\mathrm{t}+1)}^3\right)=({\mathrm{t}}^3+{3\mathrm{t}}^2+3)(9+(3+{\mathrm{C}}_6^2)\mathrm{t}+(1+{\mathrm{C}}_6^3){\mathrm{t}}^2+...+{\mathrm{t}}^5)\to 3\times 9=27$$$$\Rightarrow {\lim }_{\mathrm{x}\to 1}\mathrm{f}\left(\mathrm{x}\right)=27$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com