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  • Evaluate <img alt="\mathrm{ \lim _{x \rightarrow+\infty}\left(\frac{x^2-x+1}{x^2}\right)^{\frac{-x^2}{2 x^2-1}} }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Clim%20_%7Bx

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


 

Option: 1

\mathrm{e^{\frac{3}{2}}}


Option: 2

\mathrm{e^{\frac{5}{2}}}


Option: 3

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


Option: 4

0


Answers (1)

best_answer

Since \mathrm{e^x \: and\: \ln} are continuous functions, we obtain:

\mathrm{\lim_{x\rightarrow +\infty }\left ( \frac{x^{2}-x+1}{x^{2}} \right )^{\frac{-3x^{3}}{2x^{2}-1}}=\lim_{x\rightarrow +\infty }\left ( 1+\frac{x^{2}-x+1}{x^{2}}-1 \right )^{\frac{1}{\frac{x^{2}-x+1}{x^{2}}}-1}.\left ( \frac{x^{2}-x+1}{x^{2}}-1 \right )\frac{-3x^{3}}{2x^{2}-1}=}

\mathrm{=\lim _{x \rightarrow+\infty}\left(1+\frac{x^2-x+1}{x^2}-1\right)^{\frac{1}{\frac{x^2-x+1}{x^2}} \cdot \frac{3\left(x^2-x\right)}{2 x^2-1}}=\lim _{x \rightarrow+\infty} e^{\frac{3\left(x^2-x\right)}{2 x^2-1} \cdot \ln \left(1+\frac{x^2-x+1}{x^2}-1\right)^{\frac{1}{x^2-x+1}-1}}=}

\mathrm{=e^{\lim _{x+\infty} \frac{x\left(x^2-x\right)}{x^2-1} \cdot \ln \lim _{x \rightarrow+\infty}\left(1+\frac{x^2-x+1}{x^2}-1\right)^{\frac{1}{x^2-x+1} x^2-1}}=e^{\frac{3}{2}} .}

Hence option 1 is correct.

 

 

Posted by

Devendra Khairwa

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