Question

Consider the function $f(x)=x^{30}\cdot (\ln x{)}^{20}$ for $x>0$
If $f$ is continuous at $x=0$ then $f(0)$

• A. is equal to 0
• B. is equal to $2/3$
• C. is equal to 1
• D. can not be defined to make $f(x)$ continuous at $x=0$

Answer 1

Answer: (A)

$f(0)=\underset{x\to 0}{\mathrm{Lim}}f(x)=\underset{x\to 0}{\mathrm{Lim}}(\ln x{)}^{20}\cdot x^{30}=\underset{x\to 0}{\mathrm{Lim}}\frac{(\ln x{)}^{20}}{x^{-30}}\left(\frac{\infty }{\infty }\right)$
$=\underset{x\to 0}{\mathrm{Lim}}\frac{20(\ln x{)}^{19}}{-30\cdot x^{-31}}\frac{1}{x}=-\frac{20}{30}\frac{(\ln x{)}^{19}}{x^{-30}}\dots \dots \dots ...\text{ so on }$
hence $\underset{x\to 0}{\mathrm{Lim}}f(x)=0=f(0)\Rightarrow f$ is continuous at $x=0$ with $f(0)=0$.
$\text{ now }{f}^{\prime }(x)=x^{29}20(\ln x{)}^{19}+(\ln x{)}^{20}\cdot 30\cdot x^{29}$
$=(\ln x{)}^{19}\cdot x^{29}\cdot 10[2+3\ln x]=0$
this gives either $x=0$ which is rejected
$\text{ or }\ln x=0\Rightarrow x=1$
or $\ln x=-\frac{2}{3}\Rightarrow x=e^{-\frac{2}{3}}$