Indeterminate forms 0^0 and 1^\infty. Evaluate the following limits.

Indeterminate forms ${0}^{0}$ and ${1}^{\mathrm{\infty }}$. Evaluate the following limits.
a) $\underset{x\to {0}^{+}}{lim}{x}^{x}$
b) $\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{x}\right)}^{x}$
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Wasither1957
Given to find:
a)$\underset{x\to {0}^{+}}{lim}{\left(x\right)}^{x}$
b) $\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{x}\right)}^{x}$
Part a:
Since $u={e}^{\mathrm{ln}\left(u\right)}$, then:
$\underset{x\to {0}^{+}}{lim}{x}^{x}=\underset{x\to {0}^{+}}{lim}{e}^{\mathrm{ln}\left({x}^{x}\right)}$
$\underset{x\to {0}^{+}}{lim}{e}^{\mathrm{ln}\left({x}^{x}\right)}=\underset{x\to {0}^{+}}{lim}{e}^{x\mathrm{ln}\left(x\right)}$
Move the limit under the exponential:
$\underset{x\to {0}^{+}}{lim}{e}^{x\mathrm{ln}\left(x\right)}={e}^{\underset{x\to {0}^{+}}{lim}{e}^{x\mathrm{ln}\left(x\right)}}$
${e}^{\underset{x\to {0}^{+}}{lim}x\mathrm{ln}\left(x\right)}={e}^{\underset{x\to {0}^{+}}{lim}\frac{\mathrm{ln}\left(x\right)}{\frac{1}{x}}}$
Since we have an indeterminate form of type $\frac{\mathrm{\infty }}{\mathrm{\infty }}$, we can apply the LHospital
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juniorekze
Since $u={e}^{\mathrm{ln}\left(u\right)}$, then:
$\underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{x}\right)}^{x}=\underset{x\to \mathrm{\infty }}{lim}{e}^{\mathrm{ln}\left({\left(1+\frac{1}{x}\right)}^{x}\right)}$
$\underset{x\to \mathrm{\infty }}{lim}{e}^{\mathrm{ln}\left({\left(1+\frac{1}{x}\right)}^{x}\right)}=\underset{x\to \mathrm{\infty }}{lim}{e}^{x\mathrm{ln}\left(1+\frac{1}{x}\right)}$
$\underset{x\to \mathrm{\infty }}{lim}{e}^{x\mathrm{ln}\left(1+\frac{1}{x}\right)}={e}^{\underset{x\to \mathrm{\infty }}{lim}x\mathrm{ln}\left(1+\frac{1}{x}\right)}$
${e}^{\underset{x\to \mathrm{\infty }}{lim}x\mathrm{ln}\left(1+\frac{1}{x}\right)}={e}^{\underset{x\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(1+\frac{1}{x}\right)}{\frac{1}{x}}}$
Since we have an indeterminate form type $\frac{0}{0}$, we can apply the LHopitals rule:
${e}^{\underset{x\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(1+\frac{1}{x}\right)}{\frac{1}{x}}}={e}^{\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{d}{dx}\mathrm{ln}\left(1+\frac{1}{x}\right)}{\frac{d}{dx}\frac{1}{x}}}$
${e}^{\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{d}{dx}\mathrm{ln}\left(1+\frac{1}{x}\right)}{\frac{d}{dx}\frac{1}{x}}}={e}^{\underset{x\to \mathrm{\infty }}{lim}\frac{1}{1+\frac{1}{x}}}$
${e}^{\underset{x\to \mathrm{\infty }}{lim}\frac{1}{1+\frac{1}{x}}}={e}^{\underset{x\to \mathrm{\infty }}{lim}\frac{x}{x+1}}$