limit question: Why does numerator have to be zero for...?

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nigel

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Sep 9, 2006
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The bolded part is the main part I don't understand, please help me out, thanks in advance

lim (3x^2 + ax + a + 3) / (x^2 + x - 2)
x->-2

Question : Is there a number a such that exists? If so, find the value of a and the value of the limit.

I have a solution manual it stated: since the denominator approaches 0 as x->-2, the limit will exist only if the numerator also approaches 0 as x->-2 don't understand this can you explain it to me, i understand that the denominator is 0 but not why the numerator has to be 0 also for the limit to exist i don't get

therefore lim (3x^2 + ax + a + 3) = 0
x->-2

so (3(-2)^2 + a(-2) + a + 3) = 0

a = 15


lim (3x^2 + 15x + 18) / (x^2 + x - 2)
x->-2


then you lim (3(x+2)(x+3)) / ((x+2)(x-1))
x->-2

lim (3(x+3)) / (x-1))
x->-2

(3(-2+3)) / (-2-1)) = 3/-3 = -1

Thanks alot, Nigel
 
If \(\displaystyle f(a) = 0\;\& \;g(a) = 0\) then \(\displaystyle (x-a)\) if a factor of both f & g.
Therefore, the factor divides out and the limit could exist.
 
Re: limit question

nigel said:
The bolded part is the main part I don't understand, please help me out, thanks in advance

lim (3x^2 + ax + a + 3) / (x^2 + x - 2)
x->-2

Question : Is there a number a such that exists? If so, find the value of a and the value of the limit.

I have a solution manual it stated: since the denominator approaches 0 as x->-2, the limit will exist only if the numerator also approaches 0 as x->-2 don't understand this can you explain it to me, i understand that the denominator is 0 but not why the numerator has to be 0

With denominator = 0 and numerator \(\displaystyle *\) 0 , you will have the limit as "±infinity" - which is "undefined" for limit purposes. In that case, limit does not exist - or the function behaves (grows or shrinks) without limit(bound)

also for the limit to exist i don't get

therefore lim (3x^2 + ax + a + 3) = 0
x->-2

so (3(-2)^2 + a(-2) + a + 3) = 0

a = 15


lim (3x^2 + 15x + 18) / (x^2 + x - 2)
x->-2


then you lim (3(x+2)(x+3)) / ((x+2)(x-1))
x->-2

lim (3(x+3)) / (x-1))
x->-2

(3(-2+3)) / (-2-1)) = 3/-3 = -1

Thanks alot, Nigel
Click to expand...
 
I will try a slightly different explanation for you.

[3x^2+ax+a+3]/[x^2+x-2] factor denominator
[3x^2+ax+a+3]/[ [x+2][x-1] ]

lim x-> -2 [-6-2a+a+3]/0
lim x->0 = [-3-a]/0 infinity ,undefined if a=-3

If numerator is divisible by [x+2] with no remainder, then the pole at x=-2 is removed. Divide the numerator by x+2

If we divide we obtain 3x+[a-6] with remainder -a+15
But we want a zero remainder. Set -a+15 equal to 0
-a+15=0
a=15 answer

Proof
[3x^2+15x+18] / [ [x+2][x-1] ] factor out 3 from numerator
3[x^2+5x+6] / [ [x+2][x-1] ] factor numerator
3[x+3][x+2] /[ [x+2][x-1] ]
3[x+3] / [x-1]
lim x-> -2 3[1]/-3
lim x-> -2 f[x]=-1 answer

I hope this helps.
Arthur
 
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