Science:Math Exam Resources/Courses/MATH100/December 2019/Question 3 (b)/Solution 1

Whenever you see a variable exponent take the logarithm followed by the exponential. In this case $x^{x}=e^{xln(x)}$. Then, we want to compute the limit $\displaystyle {\lim _{x\rightarrow 0^{+}}}e^{xln(x)}.$

Notice that using L'Hopital it's possible to compute the limit $\displaystyle {\lim _{x\rightarrow 0^{+}}}xln(x)=\displaystyle {\lim _{x\rightarrow 0^{+}}}{\frac {ln(x)}{\frac {1}{x}}}=\displaystyle {\lim _{x\rightarrow 0^{+}}}{\frac {\frac {1}{x}}{-{\frac {1}{x^{2}}}}}=0.$

Therefore, since the exponential is a continuous function, we have that $\displaystyle {\lim _{x\rightarrow 0^{+}}}e^{xln(x)}=e^{\left(\displaystyle {\lim _{x\rightarrow 0^{+}}}xln(x)\right)}=e^{0}=1.$