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Question Number 80455 by jagoll last updated on 03/Feb/20

$$\underset{x\to 0}{\lim }{\left(\frac{1+mx}{1-nx}\right)}^{\frac{mn}{x}}$$

Commented by mr W last updated on 03/Feb/20

$$\underset{x\to 0}{\lim }{\left(\frac{1+mx}{1-nx}\right)}^{\frac{mn}{x}}$$$$=\underset{x\to 0}{\lim }{\left(\frac{\frac{1}{x}+m}{\frac{1}{x}-n}\right)}^{\frac{mn}{x}}$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[{\left(\frac{t+m}{t-n}\right)}^t\right]}^{mn}$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[{(1+\frac{m+n}{t-n})}^t\right]}^{mn}$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[{(1+\frac{m+n}{t-n})}^{t-n}{(1+\frac{m+n}{t-n})}^n\right]}^{mn}$$$$=\underset{t\to \mathrm{\infty }}{\lim }{\left[{(1+\frac{1}{\frac{t-n}{m+n}})}^{\frac{t-n}{m+n}}{(1+\frac{m+n}{t-n})}^{\frac{n}{m+n}}\right]}^{mn(m+n)}$$$$={\left[e{(1+0)}^{\frac{n}{m+n}}\right]}^{mn(m+n)}$$$$=e^{mn(m+n)}$$

Commented by john santu last updated on 03/Feb/20

$$\underset{x\to 0}{\lim }{(1+\frac{(n+m)x}{1-nx})}^{\frac{mn}{x}}$$$$\underset{x\to 0}{\lim }{(1+\frac{1}{\left(\frac{1-nx}{(n+m)x}\right)})}^{\frac{mn}{x}}$$$$e^{\underset{x\to 0}{\lim }\left(\frac{mn(m+n)x}{(1-nx)x}\right)}=e^{mn(m+n)}.$$

Answered by Joel578 last updated on 03/Feb/20

$$L=\underset{x\to 0}{\lim }\left\{{\left(\frac{1+mx}{1-nx}\right)}^{\frac{mn}{x}}\right\}$$$$\ln L=\underset{x\to 0}{\lim }\{\frac{mn}{x}.\ln \left(\frac{1+mx}{1-nx}\right)\}$$$$=\lim \left\{\frac{\ln (1+mx)-\ln (1-nx)}{\frac{x}{mn}}\right\}$$$$\mathrm{Using}{L}^{\prime }hopital,\mathrm{we}\mathrm{get}$$$$\ln L=\underset{x\to 0}{\lim }\left\{\frac{\frac{m}{1+mx}-\frac{(-n)}{1-nx}}{\frac{1}{mn}}\right\}=mn(m+n)$$$$\Rightarrow L=e^{mn(m+n)}$$

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