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Evaluate \mathrm{\lim _{x \rightarrow 0}\left(\frac{2+x^2}{2-x^2}\right)^{\frac{1}{z^2}}}

Option: 1

e


Option: 2

0


Option: 3

1


Option: 4

-1


Answers (1)

best_answer

                                                   \mathrm{\frac{2+x^2}{2-x^2}=1+\frac{2 x^2}{2-x^2}=1+\frac{x^2}{1-\frac{x^2}{2}}}

\mathrm{\text { and } y=\frac{x^2}{1-\frac{x^2}{2}} \rightarrow 0 \text {, as } x \rightarrow 0 \text {. Hence }}

                                                \mathrm{\left(1+\frac{x^2}{1-\frac{x^2}{2}}\right)^{\frac{1-\frac{z^2}{2}}{z^2}}=(1+y)^{1 / y} \rightarrow e}

\mathrm{as\: x\rightarrow 0}

Finally

                                 \mathrm{\left(\frac{2+x^2}{2-x^2}\right)^{1 / x^2}=\left(\left(1+\frac{x^2}{1-\frac{x^2}{2}}\right)^{\frac{1-\frac{x^2}{2}}{x^2}}\right)^{\frac{1}{1-\frac{x^2}{2}}} \rightarrow e}

\mathrm{\text { since, if } f(x) \rightarrow e \text { and } g(x) \rightarrow 1 \text {, then } f(x)^{g(z)} \rightarrow e \text {. }}

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seema garhwal

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