In our case, we have an uncertainty of type 0/0.
It follows from the fact that the denominator tends to zero when a tends to zero, and hence the numerator also tends to zero, otherwise the limit would tend to infinity.
x2+bx+4b=0 if x=a and due to factor theorem we can write:
x2+bx+4b=(x−a)(x−c) , where c is some constant. To make limit equal to -6 we should choose c=a+6 .
x→alimx−ax2+bx+4b=x→alimx−a(x−a)(x−a−6)=x→alim(x−a−6)=−6
x2+bx+4b=(x−a)(x−a−6)
x2+bx+4b=x2−x(2a+6)+a2+6a
Equating coefficients, we get:
b=−(2a+6)
4b=a2+6a
Substituting first equation into the second we will obtain:
−4(2a+6)=a2+6a
a2+14a+24=0
Using Vieta's formulas we can write
a=−12, b=−(2a+6)=−(2⋅(−12)+6)=18
a=−2, b=−(2⋅(−2)+6)=−2
Answer: a=−12 , b=18 or a=−2, b=−2 .
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