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Question Number 162834 by mkam last updated on 01/Jan/22

Answered by abdullahhhhh last updated on 01/Jan/22

$$\underset{x\to 0}{\lim }\frac{\mathbf{x}+\mathbf{tan}2\mathbf{x}}{\mathbf{x}-\mathbf{tan}2\mathbf{x}}/\mathbf{x}$$$$\underset{x\to 0}{\lim }\frac{1+2}{1-2}=-3$$$$$$

Answered by abdullahhhhh last updated on 01/Jan/22

$$\underset{x\to \mathbf{e}}{\lim }\frac{\mathbf{sinx}-\mathbf{x}}{{\mathbf{x}}^3}=\frac{\mathbf{sin}\left(\mathbf{e}\right)-\mathbf{e}}{{\mathbf{e}}^3}$$$$$$

Answered by abdullahhhhh last updated on 01/Jan/22

$$\underset{x\to 0}{\lim }\frac{{\mathbf{e}}^{\mathbf{x}}+{\mathbf{e}}^{-\mathbf{x}}-{\mathbf{x}}^2-2}{{\mathbf{sin}}^2\mathbf{x}-{\mathbf{x}}^2}=\left(\frac{0}{0}\right)\mathbf{use}\mathbf{lobital}$$$$\underset{x\to 0}{\lim }\frac{{\mathbf{e}}^{\mathbf{x}}-{\mathbf{e}}^{-\mathbf{x}}-2\mathbf{x}}{2\mathbf{sinx}\mathbf{cosx}\left(\mathbf{sin}2\mathbf{x}\right)-2\mathbf{x}}=\frac{0}{0}$$$$\underset{x\to 0}{\lim }\frac{{\mathbf{e}}^{\mathbf{x}}+{\mathbf{e}}^{-\mathbf{x}}-2}{2\mathbf{cos}2\mathbf{x}-2}=\frac{2-2}{2-2}=\frac{0}{0}$$$$\underset{x\to 0}{\lim }\frac{{\mathbf{e}}^{\mathbf{x}}-{\mathbf{e}}^{-\mathbf{x}}}{-4\mathbf{sin}2\mathbf{x}}=\frac{1-1}{0}=\frac{0}{0}$$$$\underset{x\to 0}{\lim }\frac{{\mathbf{e}}^{\mathbf{x}}+{\mathbf{e}}^{-\mathbf{x}}}{-8\mathbf{cos}2\mathbf{x}}=\frac{2}{-8}=\frac{-1}{4}$$

Answered by abdullahhhhh last updated on 01/Jan/22

$$\underset{x\to 0}{\lim }\frac{{\mathbf{xe}}^{\mathbf{x}}-\mathbf{ln}(1+\mathbf{x})}{{\mathbf{x}}^2}=\frac{0}{0}\mathbf{lobital}$$$$\underset{x\to 0}{\lim }\frac{{\mathbf{xe}}^{\mathbf{x}}+{\mathbf{e}}^{\mathbf{x}}-\frac{1}{(\mathbf{x}+1)}}{2\mathbf{x}}=\frac{0}{0}$$$$\underset{x\to 0}{\lim }\frac{{\mathbf{xe}}^{\mathbf{x}}+{\mathbf{e}}^{\mathbf{x}}+{\mathbf{e}}^{\mathbf{x}}+\frac{1}{{(\mathbf{x}+1)}^2}}{2}=\frac{3}{2}$$

Answered by abdullahhhhh last updated on 01/Jan/22

$$\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{\mathbf{lnx}}{\mathbf{x}}\right)=\frac{\mathrm{\infty }}{\mathrm{\infty }}\mathbf{use}\mathbf{lobital}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\mathbf{x}}=\frac{1}{\mathrm{\infty }}=0$$

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