Question

is this right??

x^-1/2 -1/x^1/2 all over 3x^-1/3 -3/x^1/3

=1/x^1/2 -1/x^1/2 all over 3/x^1/3 - 3/x^1/3

=O? thanks guys :) owe you a lot

Answer 1

yep! As in:\[x^{-a} = \frac{ 1 }{ x^{a} }\]

Answer 2

@curiousshubham \[x- _{2}^{1} -1/x _{2}^{1} \over 3x- _{3}^{1} -3/x _{3}^{1}\]thankyou

Answer 3

you are welcomed @xtel .

Answer 4

@curiousshubham theres the question ^^

Answer 5

is the numerator this :
\(\huge x^{-1/2}-\frac{1}{x^{1/2}}\)

or this :

\(\huge \frac{x^{-1/2}-{1}}{x^{1/2}}\)

Answer 6

If its first one then the solution is indeterminate.

Answer 7

yup.

Answer 8

@halcy0n @curiousshubham yes the first one.

it could be turn into this to make it positive? \[1/x _{2}^{1} -1/x _{2}^{1}\]

Answer 9

yes, so u get 0 in numerator.
but similarly, then u also get 0 in denominator
so u have 0/0 which is an indeterminate form

Answer 10

@hartnn the first one :)

Answer 11

@xtel So the solution is indeterminate.

Answer 12

@curiousshubham @hartnn thankyou again guys :)) i owe you a lot