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Question Number 154776 by john_santu last updated on 21/Sep/21

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

Commented by mathdanisur last updated on 21/Sep/21

$$\mathrm{say}\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow \mathrm{t}\to 0$$$${\mathrm{e}}^{\underset{\mathbf{t}\to 0}{\lim }\left(\frac{\mathbf{t}-\mathbf{sin}\left(\mathbf{t}\right)}{\mathbf{t}(\mathbf{t}+1)}\right)\frac{1}{\mathbf{t}}}={\mathrm{e}}^{\frac{1}{\mathbf{e}}\cdot \mathbf{t}\to 0\left(\frac{1-\mathbf{cos}\left(\mathbf{t}\right)}{2\mathbf{t}}\right)}={\mathrm{e}}^0=1$$

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