Question

find limit as x approaches - infinity
find limit as x approaches infinity

y= 9x ln|x|

Answer 1

are you familiar with L hospital rule

Answer 2

yes. oh
would you take the derivative of 9x over derivative of ln|x|

Answer 3

tryit out :)

Answer 4

y'=9 ln(x) + 9x/x , where x >0 as log is defined for +ve values
then evaluaate limits , if you still encounter infinity, take y" = 9/x, evaluate limits and answer is 9/ infinity =0

Answer 5

wait, why did you divide by x. i thought it was f(x)/g'(x) so derivative of 9x/ derivative of lnx. and why is 9lnx added with 9x

Answer 6

derivative of ln(x) = 1/x

Answer 7

product rule

Answer 8

oh! thanks

Answer 9

wait, there's supposed to be two answers since it's the limit approacing negative and postive infinity. 0 isn't an answer

Answer 10

whats the answer

Answer 11

ok then this needs a trick here it is

write 9xln(x) =\[9 \ln (x)/\frac{ 1 }{ x }\], which is essentialy same thing

observer it is infinity over 0 form

use hopitals rule

9 (1/x) is derivative of numerator

-1/x^2 is that of denominator

on cancellation of common terms you will have -9x in numerator,

plug value of x = infinity, you get your answer