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Question Number 87153 by jagoll last updated on 03/Apr/20

$$\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}-2}{{\mathrm{x}}^2}$$

Commented by john santu last updated on 03/Apr/20

$$\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{2\mathrm{x}}=\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}=1$$

Commented by mathmax by abdo last updated on 03/Apr/20

$$wehave{e}^x=1+\frac{x}{1!}+\frac{x^2}{2!}+o\left(x^3\right)(x\to 0)$$$$e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}+o\left(x^3\right)\Rightarrow e^x+e^{-x}-2=x^2+o\left(x^3\right)\Rightarrow$$$$\frac{e^x+e^{-x}-1}{x^2}=1+o\left(x\right)\Rightarrow {lim}_{x\to 0}\frac{e^x+e^{-x}-2}{x^2}=1$$

Answered by $@ty@m123 last updated on 03/Apr/20

$$\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}+\frac{1}{e^x}-2}{{\mathrm{x}}^2}$$$$\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{2x}+1-2e^x}{{\mathrm{x}}^2.e^x}$$$$\underset{x\to 0}{\lim }\frac{{(e^x-1)}^2}{{\mathrm{x}}^2.e^x}$$$$\underset{x\to 0}{\lim }{\left(\frac{e^x-1}{x}\right)}^2\times \frac{1}{\underset{x\to 0}{\lim }{e}^x}$$$$=1^2\times 1$$$$=1$$

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