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Question Number 112863 by bemath last updated on 10/Sep/20

$$\left(1\right)\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{2\mathrm{x}+3}{2\mathrm{x}-1}\right)}^{4\mathrm{x}+2}$$$$\left(2\right)\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{3\mathrm{x}+1}{3\mathrm{x}-1}\right)}^{4\mathrm{x}-2}$$

Commented by bobhans last updated on 22/Sep/20

$$\left(1\right)\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{2x+3}{2x-1}\right)}^{4x+2}=e^{\underset{x\to \mathrm{\infty }}{\lim }(\frac{2x+3}{2x-1}-1).(4x+2)}$$$$=e^{\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{4(4x+2)}{2x-1}\right)}=e^8$$

Answered by bobhans last updated on 10/Sep/20

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{3\mathrm{x}-1+2}{3\mathrm{x}-1}\right)}^{4\mathrm{x}-2}=\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{2}{3\mathrm{x}-1})}^{4\mathrm{x}-2}$$$$\mathrm{let}\mathrm{u}=\frac{2}{3\mathrm{x}-1},\mathrm{u}\to 0\mathrm{and}3\mathrm{x}-1=\frac{2}{\mathrm{u}}$$$$\mathrm{x}=\frac{2+\mathrm{u}}{3\mathrm{u}}.\mathrm{then}\underset{\mathrm{u}\to 0}{\lim }{(1+\mathrm{u})}^{\frac{4\mathrm{u}+8}{3\mathrm{u}}-2}$$$$\underset{\mathrm{u}\to 0}{\lim }{\left[{(1+\mathrm{u})}^{\frac{1}{\mathrm{u}}}\right]}^{\frac{8-2\mathrm{u}}{3}}={\mathrm{e}}^{\underset{\mathrm{u}\to 0}{\lim }\left(\frac{8-2\mathrm{u}}{3}\right)}={\mathrm{e}}^{\frac{8}{3}}$$

Answered by bobhans last updated on 10/Sep/20

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{2\mathrm{x}-1+4}{2\mathrm{x}-1}\right)}^{4\mathrm{x}+2}=\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{4}{2\mathrm{x}-1})}^{4\mathrm{x}+2}$$$$\mathrm{let}\mathrm{q}=\frac{4}{2\mathrm{x}-1}\to \mathrm{x}=\frac{4+\mathrm{q}}{2\mathrm{q}}$$$$\underset{\mathrm{q}\to 0}{\lim }\left[{(1+\mathrm{q})}^{\frac{8+2\mathrm{q}}{\mathrm{q}}+2}\right]=\underset{\mathrm{q}\to 0}{\lim }{\left[{(1+\mathrm{q})}^{\frac{1}{\mathrm{q}}}\right]}^{4\mathrm{q}+8}$$$$={\mathrm{e}}^{\underset{\mathrm{q}\to 0}{\lim }(4\mathrm{q}+8)}={\mathrm{e}}^8$$

Answered by mathmax by abdo last updated on 10/Sep/20

$$1)\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\left(\frac{2\mathrm{x}+3}{2\mathrm{x}-1}\right)}^{4\mathrm{x}+2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{(4\mathrm{x}+2)\ln \left(\frac{2\mathrm{x}+3}{2\mathrm{x}-1}\right)}$$$$\mathrm{we}\mathrm{have}\ln \left(\frac{2\mathrm{x}+3}{2\mathrm{x}-1}\right)=\ln \left(\frac{2\mathrm{x}-1+4}{2\mathrm{x}-1}\right)=\ln (1+\frac{4}{2\mathrm{x}-1})\sim \frac{4}{2\mathrm{x}-1}$$$$(4\mathrm{x}+2)\ln \left(\frac{2\mathrm{x}+3}{2\mathrm{x}-1}\right)\sim 4.\frac{4\mathrm{x}+2}{2\mathrm{x}-1}\sim 8\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim {\mathrm{e}}^8\Rightarrow$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^8$$

Answered by mathmax by abdo last updated on 10/Sep/20

$$2)\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)={\left(\frac{3\mathrm{x}+1}{3\mathrm{x}-1}\right)}^{4\mathrm{x}-2}\Rightarrow \mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^{(4\mathrm{x}-2)\ln \left(\frac{3\mathrm{x}+1}{3\mathrm{x}-1}\right)}$$$$\mathrm{we}\mathrm{have}\ln \left(\frac{3\mathrm{x}+1}{3\mathrm{x}-1}\right)=\ln \left(\frac{3\mathrm{x}-1+2}{3\mathrm{x}-1}\right)=\ln (1+\frac{2}{3\mathrm{x}-1})\sim \frac{2}{3\mathrm{x}-1}\Rightarrow$$$$(4\mathrm{x}-2)\ln \left(\frac{3\mathrm{x}+1}{3\mathrm{x}-1}\right)\sim 2\times \frac{4\mathrm{x}-2}{3\mathrm{x}-1}\sim \frac{8}{3}\Rightarrow {\lim }_{\mathrm{x}\to \mathrm{\infty }}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{8}{3}}$$

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