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Hello can someone help me with this, i do two times Lopital, and get 0, I am not sure is that ok?

 

\(\lim _{x\to \infty \:}\left(\left(\ln \left(x\right)+1\right)\div \:e^{\sqrt{\ln \left(x\right)+1}}\right)\)

 Sep 13, 2019
 #1
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Hello can someone help me with this, i do two times  L'Hospital,
and get \(0\), I am not sure is that ok?

\(\lim \limits_{x\to \infty}\left( \dfrac{ \left(\ln(x)+1\right) } {e^{\sqrt{\ln(x)+1}} } \right)\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \lim \limits_{x\to \infty}\left( \dfrac{ \ln(x)+1 } {e^{\sqrt{\ln(x)+1}} } \right) } \\\\ &\overset{\text{L'Hospital}}{=}& \lim \limits_{x\to \infty}\left( \dfrac{1}{x} \above 1pt \dfrac{e^{\sqrt{\ln(x)+1}} } { 2x\sqrt{\ln(x)+1} } \right) \\\\ & = & \lim \limits_{x\to \infty}\left( \dfrac{2x\sqrt{\ln(x)+1}}{x} \above 1pt e^{\sqrt{\ln(x)+1}} \right) \\\\ & = & \lim \limits_{x\to \infty}\left( \dfrac{2 \sqrt{\ln(x)+1}} { e^{\sqrt{\ln(x)+1}} } \right) \\\\ &\overset{\text{L'Hospital}}{=}& \lim \limits_{x\to \infty}\left( \dfrac{2} {2x\sqrt{\ln(x)+1} \dfrac{ e^{\sqrt{\ln(x)+1}}}{2x\sqrt{\ln(x)+1}} } \right) \\\\ & = & \lim \limits_{x\to \infty}\left( \dfrac{2} { e^{\sqrt{\ln(x)+1}} } \right) \\\\ & = & \dfrac{2}{\infty} \\\\ &=& \mathbf{0} \\ \hline \end{array}\)

 

laugh

 Sep 13, 2019

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