Using the definition and continuity of the exponential,
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Anonymous says:
This exercise can be solved without L’Hôpital’s rule putting e(1+logx-1) in the numerator, where e(1+logx)/(1+logx)=e and the limit when x approaches 0 by the right of e/(1+logx)=0.
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![\[ \lim_{x \to 0^+} x^{\frac{e}{1 + \log x}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-138181e23113d1076a019cec57c435a5_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \lim_{x \to 0^+} x^{\frac{e}{1 + \log x}} &= \lim_{x \to 0^+} e^{ \frac{e}{1+\log x} \log x } \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{e \log x}{1 + \log x} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{\frac{e}{x}}{\frac{1}{x}} \right) &(\text{L'Hopital}) \\[9pt] &= \exp \left( \lim_{x \to 0^+} e \right) \\[9pt] &= e^e. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-cba7d321699673f48a3e57c50106bf2e_l3.png "Rendered by QuickLaTeX.com")
This exercise can be solved without L’Hôpital’s rule putting e(1+logx-1) in the numerator, where e(1+logx)/(1+logx)=e and the limit when x approaches 0 by the right of e/(1+logx)=0.