Solved

A) Graph the Funtion f(x)=x2lnxf(x)=x^{2} \ln x B) Use L'Hospitals' Rule to Explain the Behavior As L'Hospitals

Question 151

Multiple Choice

a) Graph the funtion f(x) =x2lnxf(x) =x^{2} \ln x . b) Use l'Hospitals' rule to explain the behavior as x0x \rightarrow 0


A) a)  a)  Graph the funtion  f(x) =x^{2} \ln x  . b)  Use l'Hospitals' rule to explain the behavior as  x \rightarrow 0  A)  a)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=0  There is a hole at (0, 0)  B)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=1  There is a hole at (0, 1)  C)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=1  There is a hole at (1, 0)  D)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=0  There is a hole at (0, 0)  b) limx0+x2lnx=0\lim _{x \rightarrow 0_{+}} x^{2} \ln x=0
There is a hole at (0, 0)
B)  a)  Graph the funtion  f(x) =x^{2} \ln x  . b)  Use l'Hospitals' rule to explain the behavior as  x \rightarrow 0  A)  a)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=0  There is a hole at (0, 0)  B)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=1  There is a hole at (0, 1)  C)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=1  There is a hole at (1, 0)  D)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=0  There is a hole at (0, 0)  b) limx0+x2lnx=1\lim _{x \rightarrow 0_{+}} x^{2} \ln x=1
There is a hole at (0, 1)
C)  a)  Graph the funtion  f(x) =x^{2} \ln x  . b)  Use l'Hospitals' rule to explain the behavior as  x \rightarrow 0  A)  a)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=0  There is a hole at (0, 0)  B)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=1  There is a hole at (0, 1)  C)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=1  There is a hole at (1, 0)  D)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=0  There is a hole at (0, 0)  b) limx0+x2lnx=1\lim _{x \rightarrow 0_{+}} x^{2} \ln x=1
There is a hole at (1, 0)
D)  a)  Graph the funtion  f(x) =x^{2} \ln x  . b)  Use l'Hospitals' rule to explain the behavior as  x \rightarrow 0  A)  a)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=0  There is a hole at (0, 0)  B)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=1  There is a hole at (0, 1)  C)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=1  There is a hole at (1, 0)  D)    b)   \lim _{x \rightarrow 0_{+}} x^{2} \ln x=0  There is a hole at (0, 0)  b) limx0+x2lnx=0\lim _{x \rightarrow 0_{+}} x^{2} \ln x=0
There is a hole at (0, 0)

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions