Question

lim x→0+[x√9][lnx]

Answer 1

\[\lim_{x \rightarrow 0^+}\frac{x}{\ln x}=0\]
Using L'Hopital's rule

Answer 2

Could you should the steps..

Answer 3

show*

Answer 4

Shoot sorry i wrote the problem wrong.. it should be \[\sqrt[9]{x}\]

Answer 5

times lnx

Answer 6

ok then it becomes
\[\lim_{x \rightarrow 0+} \frac{\ln(x)}{x^{-1/9}} =\lim_{x \rightarrow 0+}\frac{1/x}{(-1/9)x^{-10/9}} =\lim_{x \rightarrow 0+}-9x^{1/9} =0\]

Answer 7

i don't understand why you're putting ln (x) over x^-1/9

Answer 8

\[x^{1/9} = \frac{1}{x^{-1/9}}\]

Answer 9

its to put the function in a form where you can apply L'Hopitals rule, there needs to be a function of x in numerator and denominator

Answer 10

can you show in detail how you got -9x^1/9?

Answer 11

oh sure it comes from multiplying by the reciprocal of bottom fraction
\[\large \frac{\frac{1}{x}}{-\frac{x^{-10/9}}{9}} = \frac{1}{x}*-\frac{9}{x^{-10/9}} = -\frac{9}{x^{-1/9}} =-9x^{1/9}\]