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

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

Option: 1

4


Option: 2

3


Option: 3

-2


Option: 4

1  (correct) 


Answers (1)

best_answer

We can use exponential trick in this way

                                \mathrm{ \left(\frac{x^2}{2 x+1}\right)^{1 / x}=e^{\frac{\log \left(\frac{z^2}{2 x+1}\right)}{z}} \rightarrow e^0=1 }

indeed by standard limit \mathrm{\frac{\log x}{x} \rightarrow 0} we have

                               \mathrm{ \frac{\log \left(\frac{x^2}{2 x+1}\right)}{x}=\frac{\log \left(\frac{x^2}{2 x+1}\right)}{\frac{x^2}{2 x+1}} \frac{x}{2 x+1} \rightarrow 0 \cdot \frac{1}{2}=0 }

Posted by

Devendra Khairwa

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