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Question Number 127459 by liberty last updated on 30/Dec/20

$$\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x+\frac{1}{2+\frac{1}{x}}})}^{x^2}=?$$

Answered by john_santu last updated on 30/Dec/20

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

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