How do you find the limit of #(3x^2+6x)/(x^2-4)# as x approaches 2?

Question

How do you find the limit of  #(3x^2+6x)/(x^2-4)# as x approaches 2?

Paisley Johnson · Accepted Answer

Please see below !

#3x^2+6x=3x(x+2)# #x^2-4=(x-2)(x+2)# #(3xcancel((x+2)))/((x-2)cancel((x+2)))# #(3x)/(x-2)# I think you have Wrote the question wrongly and it approaches to #(-2) # instead Or if it as you said approaches to 2 So, the answer is D.N.E (doesn't exist look at the graph the limit for the 2 from the left is #-oo# and from right #oo# SO IT DOEN't exist. graph{(3x^2+x6)/(x^2-4) [-8.38, 11.62, -9.28, 0.72]} ) BUT if it is approaches to -2 it gice you a value (#(3x)/(x-2)#) BLug in here,

Christopher Agresta · Answer

undefined To find the limit of (3x^2+6x)/(x^2-4) as x approaches 2, we can substitute the value of 2 into the expression. However, this would result in division by zero, which is undefined. Therefore, we need to simplify the expression by factoring the denominator. The denominator can be factored as (x+2)(x-2). Canceling out the common factor of (x-2) in the numerator and denominator, we are left with 3x/(x+2). Now, substituting x=2 into this simplified expression, we get 3(2)/(2+2) = 6/4 = 3/2. Therefore, the limit of (3x^2+6x)/(x^2-4) as x approaches 2 is 3/2.