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Question Number 138959 by mathlove last updated on 20/Apr/21

Answered by mathmax by abdo last updated on 20/Apr/21

$${\mathrm{a}}^{{\mathrm{x}}^2-1}={\mathrm{e}}^{({\mathrm{x}}^2-1)\mathrm{loga}}\sim 1+({\mathrm{x}}^2-1)\mathrm{loga}\mathrm{and}$$$${\mathrm{b}}^{{\mathrm{x}}^3-1}={\mathrm{e}}^{({\mathrm{x}}^3-1)\mathrm{logb}}\sim 1+({\mathrm{x}}^3-1)\mathrm{logb}$$$${\mathrm{c}}^{{\mathrm{x}}^5-1}\sim {\mathrm{e}}^{({\mathrm{x}}^5-1)\mathrm{logc}}\sim 1+({\mathrm{x}}^5-1)\mathrm{logc}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{a}}^{{\mathrm{x}}^2-1}-{\mathrm{b}}^{{\mathrm{x}}^3-1}}{{\mathrm{c}}^{{\mathrm{x}}^5-1}-1}\sim \frac{1+({\mathrm{x}}^2-1)\mathrm{loga}-1-({\mathrm{x}}^3-1)\mathrm{logb}}{({\mathrm{x}}^5-1)\mathrm{logc}}$$$$\sim \frac{(\mathrm{x}+1)\mathrm{loga}-({\mathrm{x}}^2+\mathrm{x}+1)\mathrm{logb}}{(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4)\mathrm{logc}}\Rightarrow$$$${\lim }_{\mathrm{x}\to 1}\mathrm{f}\left(\mathrm{x}\right)=\frac{2\mathrm{loga}-3\mathrm{logb}}{5\mathrm{logc}}$$

Answered by mitica last updated on 21/Apr/21

$$\underset{x\to }{lim}\frac{\frac{a^{x^2-1}-1}{x^2-1}(x-1)(x+1)-\frac{b^{x^3-1}-1}{x^3-1}(x-1)(x^2+x+1)}{\frac{c^{x^5-1}-1}{x^5-1}(x-1)(x^4+x^3+x^2+x+1)}=$$$$=\frac{2lna-3lnb}{5lnc}$$

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