How do you find the limit of #(-2x^2+x)/x# as x approaches 0?
1 Answer
May 21, 2016
1
Explanation:
If we substitute x = 0 into the function we obtain
#0/0#
which is indeterminate.However, factorising the numerator gives:
#(x(-2x+1))/x=(cancel(x) (-2x+1))/cancel(x)=-2x+1# now
#lim_(xto 0)(-2x+1)=0+1=1#
#rArrlim_(xto 0)(-2x^2+x)/x=1#
