Calculate the limit of the function, if it exists? limit (x^2+x-6)/(x-2), x->2
lim (x^2+x-6)/(x-2) when x-->2
first we substitute with x=2
==> lim 2^2+2-6/2-2= 0/0.... this method failed, because 2 is the root for the upper and lower functions. That means, x-2 is a common factor. So, we need to factorize upper and lower function to get rid of the common factor in order to find the limit.
==> lim (X^2+X-6)/(X-2)= lim (x-2)(x+3)/(x-2)
==> lim (x+3) when x--> 2 = lim (2+3) = 5
Lt (x^2+x-6)/(x-2) as x-->2.
As x-->2, numerator = x^2+x-6 = 0 and denominator (x-2) = 2-2 = 0. The limit as x-2, appears like 0/0 form or indeterminate for.So by remainder theorem, numarator is factor of x-2. So we can reduce both numerator and denominator by x-2 and then take limit.
On the otherhand , we can L'Hospital's method of diffrentiating numerator and denominator and then allowing x--> 2. :
Lt(x^2+x-6)/(x-2) as x-->2 is equal to:
Lt (x^2+x-6)'/(x-2)' as x-->2= Lt(2x+1)/1 as x-->2 = (2*2+1)/1 = 5.
We couldn't calculate the lim, by substituting x=2, because f(2) is not defined.
We cannot apply the Quotient Law, also, because the limit of denominator is 0, too.
We'll factorize the numerator, after finding it's roots
We'll use the quadratic formula:
After reducing the terms:
lim (x^2+x-6)/(x-2)=lim (x+3) = 2+3 =5