# Calculate the limit of the function, if it exists? limit (x^2+x-6)/(x-2), x->2

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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.

Solution:

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

(x^2+x-6)/(x-2)

x^2+x-6=0

We'll use the quadratic formula:

x1=[-1+sqrt(1+24)]/2

x1=(-1+5)/2

x1=2

x2=(-1-5)/2

x2=-3

(x^2+x-6)/(x-2)=(x-2)(x+3)/(x-2)

After reducing the terms:

**lim (x^2+x-6)/(x-2)=lim (x+3) = 2+3 =5**