How do you evaluate #lim_(x->3)(2x^2-5x-3)/(x-3)#?

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Answer 1 · 2017-11-08T13:51:58

#lim_(xrarr3)(2x^2-5x-3)/(x-3) = lim_(xrarr3)((x-3)(2x+1))/(x-3) = lim_(xrarr3)(2x+1) = 2(3)+1 = 7#

The numerator is #0# when #x=3# (#3# is a zero of the numerator).

Therefore #x-3# is a factor of the numerator. (So the quotient can be simplified for #x != 3#)

#2x^2-5x-3 = (x-3)( "something")#

By trial and error (or no error, just trial):

#2x^2-5x-3 = (x-3)(2x+1)#

So #(2x^2-5x-3)/(x-3) = ((x-3)(2x+1))/(x-3) = 2x+1# for #x != 3#.

The limit as #xrarr3# doesn't care what happens right at #3#, just for #x# close to #3#, so we can find the limit.

#lim_(xrarr3)(2x^2-5x-3)/(x-3) = lim_(xrarr3)((x-3)(2x+1))/(x-3) = lim_(xrarr3)(2x+1) = 2(3)+1 = 7#

Answer 2 · 2017-11-08T13:57:19

The function approaches 7

To find the limit of a function, we use L'Hopital's rule, where: #lim_(x->a)f(x)/g(x)=(f'(x))/(g'(x)), a!=\oo# then substitute #a# for #x#.

So, for our example: #lim_(x->3)(2x^2-5x-3)/(x-3)#, to find the limit we differentiate both equations to get:
#lim_(x->3)(4x-5)/1#

Now we replace #x# with 3:
#lim_(x->3)(4(3)-5)/1=7#