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)

$f(0) =\underset{x \rightarrow 0}{\text{Lim}} f(x)=\underset{x \rightarrow 0}{\text{Lim}}(\ln x)^{20} \cdot x^{30}=\underset{x \rightarrow 0}{\text{Lim}}\frac{(\ln x)^{20}}{x^{-30}}\left(\frac{\infty}{\infty}\right)$
$=\underset{x \rightarrow 0}{\text{Lim}} \frac{20(\ln x)^{19}}{-30 \cdot x^{-31}} \frac{1}{x}=-\frac{20}{30} \frac{(\ln x)^{19}}{x^{-30}} \ldots \ldots \ldots . . . \text { so on }$
hence $\underset{x \rightarrow 0}{\text{Lim}}f ( x )=0= f (0) \Rightarrow f$ 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}}$