A helpful procedure would be to divide by the highest denominator power.
But First, let's simplify the function given
#--------------------#
#(1/x-10x)/(2x+1/(3x))=>((1-10x^2)/x)/((6x^2+1)/(3x))=>(1-10x^2)/x*(3x)/(6x^2+1)=(3cancelx(1-10x^2))/(cancelx(6x^2+1))=(3-30x^2)/(6x^2+1)#
#--------------------#
So now we have #lim_(x->oo)(3-30x^2)/(6x^2+1)#
#x^2# is the highest denominator power so we can divide #lim_(x->oo)(3-30x^2)/(6x^2+1)# by #x^2# or multiply it by #1/x^2#. It's the same thing.
#lim_(x->oo)(3-30x^2)/(6x^2+1)*(1/x^2)/(1/x^2)=>lim_(x->)(3/x^2-(30x^2)/x^2)/((6x^2)/x^2+1/(x^2))#
#=>lim_(x->oo)(3/x^2-30)/(6+1/(x^2)#
Plugging in #oo#
#=(3/color(red)oo^2-30)/(6+1/(color(red)oo^2))#
#=(0-30)/(6+0)#
#=-30/6#
#=-5#
#:.lim_(x->oo)(1/x-10x)/(2x+1/(3x))=-5#
