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Question Number 128185 by liberty last updated on 05/Jan/21

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

Answered by john_santu last updated on 05/Jan/21

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{x-1+4}{x-1}\right)}^{x+3}=\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{4}{x-1})}^{x+3}$$$$=\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{4}{x-1})}^{(x-1)+4}$$$$=\underset{h\to \mathrm{\infty }}{\lim }{(1+\frac{4}{h})}^{h+4}$$$$=\underset{h\to \mathrm{\infty }}{\lim }{(1+\frac{4}{h})}^h.\underset{h\to \mathrm{\infty }}{\lim }{(1+\frac{4}{h})}^4$$$$=e^4.1=e^4$$

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