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  • Evaluate    <img alt="\mathrm{L=\lim _{x \rightarrow e}(2-\ln x)^{\frac{1}{z^2-x^2}}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BL%3D%5Clim%20_%7Bx%20%5Crighta

Evaluate    \mathrm{L=\lim _{x \rightarrow e}(2-\ln x)^{\frac{1}{z^2-x^2}}}

Option: 1

\mathrm{-\frac{1}{2 e^2}}


Option: 2

\mathrm{-\frac{1}{ e^2}}


Option: 3

0


Option: 4

1


Answers (1)

best_answer

                                        \mathrm{\lim _{x \rightarrow e}(2-\ln x)^{\frac{1}{x^2-e^2}}=\lim _{y \rightarrow 0}(2-\ln (y+e))^{\frac{1}{3(y+2 x)}}}

and

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

indeed

            \mathrm{\frac{1-\ln (y+e)}{y(y+2 e)}=\frac{1-\ln e-\ln (1+y / e)}{y(y+2 e)}=-\frac{1}{e} \frac{\ln (1+y / e)}{y / e} \frac{1}{(y+2 e)} \rightarrow-\frac{1}{2 e^2}}

 

Posted by

Pankaj Sanodiya

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