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Question Number 164138 by LEKOUMA last updated on 14/Jan/22

Answered by floor(10²Eta[1]) last updated on 14/Jan/22

$$\mathrm{L}=\lim (...)$$$$\mathrm{lnL}=\lim \left(\frac{\mathrm{x}+1}{{\mathrm{x}}^2+1}\right).\ln \left(\lim \left(\frac{{\mathrm{x}}^3+5}{{\mathrm{x}}^2+2}\right)\right)$$$$=\frac{\lim \left(\ln \left(\frac{{\mathrm{x}}^3+5}{{\mathrm{x}}^2+2}\right)\right)}{\frac{1}{\lim \left(\frac{\mathrm{x}+1}{{\mathrm{x}}^2+1}\right)}}=\frac{\lim \left(\ln \left(\frac{{\mathrm{x}}^3+5}{{\mathrm{x}}^2+1}\right)\right)}{\lim \left(\frac{{\mathrm{x}}^2+1}{\mathrm{x}+1}\right)}$$$$\frac{\lim \left(\frac{{3\mathrm{x}}^2({\mathrm{x}}^2+1)-2\mathrm{x}({\mathrm{x}}^3+5)}{({\mathrm{x}}^2+1)({\mathrm{x}}^3+5)}\right)}{\lim \left(\frac{2\mathrm{x}(\mathrm{x}+1)-({\mathrm{x}}^2+1)}{{(\mathrm{x}+1)}^2}\right)}$$$$\frac{\lim \left(\frac{{\mathrm{x}}^4+{3\mathrm{x}}^2-10\mathrm{x}}{{\mathrm{x}}^5+{\mathrm{x}}^3+{\mathrm{x}}^2+5}\right)}{\lim \left(\frac{{\mathrm{x}}^2+2\mathrm{x}-1}{{\mathrm{x}}^2+2\mathrm{x}+1}\right)}=\frac{\lim \left(\frac{{\mathrm{x}}^4(1+\frac{3}{{\mathrm{x}}^2}-\frac{10}{{\mathrm{x}}^3})}{{\mathrm{x}}^5(1+\frac{1}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{x}}^3}+\frac{5}{{\mathrm{x}}^5})}\right)}{\lim \left(\frac{{\mathrm{x}}^2(1+\frac{2}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^2})}{{\mathrm{x}}^2(1+\frac{2}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}}\right)}$$$$=\frac{\lim \left(\frac{{\mathrm{x}}^4}{{\mathrm{x}}^5}\right)}{\lim \left(\frac{{\mathrm{x}}^2}{{\mathrm{x}}^2}\right)}=\frac{\lim \left(\frac{1}{\mathrm{x}}\right)}{1}=\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{1}{\mathrm{x}}\right)=0$$

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