To do this, factor the numerators and denominators. Notice that we can't plug in #-2# now, because this gives us the ratio #0//0#.
To factor #3x^2+5x-2#, we look for two numbers whose product is #-6# and sum is #5#, which are #6# and #-1#.
Then, #3x^2+5x-2=3x^2+6x-x-2#
#color(white)(lll)=3x(x+2)-1(x+2)#
#color(white)(lll)=(3x-1)(x+2)#
In factoring #x^2-3x-10#, we look for the numbers whose product is #-10# and sum is #-3#, which are #-5# and #2#.
Then, #x^2-3x-10=(x-5)(x+2)#
So #lim_(xrarr-2)(3x^2+5x-2)/(x^2-3x-10)=lim_(xrarr-2)((3x-1)(x+2))/((x-5)(x+2))#, and we see that we can cancel the #(x+2)# terms, so the limit equals #lim_(xrarr-2)(3x-1)/(x-5)#.
Now plugging in #-2# won't create the issue it did before and we can plug in #x=-2#, so: #lim_(xrarr-2)(3x-1)/(x-5)=(3(-2)-1)/(-2-5)=(-7)/(-7)=1#.
