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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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


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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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