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Question Number 15186 by Joel577 last updated on 08/Jun/17

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

Commented by mrW1 last updated on 08/Jun/17

$$\mathrm{yes},\mathrm{correct}.$$

Commented by Joel577 last updated on 08/Jun/17

$$thankyouverymuch$$

Commented by Joel577 last updated on 08/Jun/17

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

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