Given that $$ \lim_{x \rightarrow 3}\frac{4x^3-bx^2+2x+30}{x^3-ax+3a-27} = \frac{1}{2} $$ Find the value of a and b. I have using $$ \lim_{x \rightarrow 3}4x^3-bx^2+2x+30 = 0 $$ To found that b = 16
Does this mean that $$ \lim_{x \rightarrow 3}\frac{4x^3-16x^2+2x+30}{x^3-ax+3a-27} = \frac{1}{2} $$
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Hint. Now that you have $b=16$, notice there is a common factor between numerator and denominator that you can cancel, by the factor theorem...
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You correctly realise that it must be a $0/0$ form since denominator goes zero for the limit, so numerator and denominator must have a common root. Since denominator has a root $x = 3$, numerator must also be divisible by $x-3$.
But as Hagan says, you don't know the multiplicity of the root $x=3$ in numerator or denominator.
If there is an extra $(x-3)$ in numerator, limit will be zero
If there is an extra $(x-3)$ in denominator, limit will not exist ($\pm \infty$ from different sides)
So we seek to factorise in such a way that multiplicity of root $x=3$ is same on numerator and denominator.
$$\frac{(x-3)(4x^2+(12-b)x+(2+3(12-b)))}{(x-3)(x^2+3x+9-a)}$$
Now in the numerator, the constant term $-3(2+3(12-b))$ must be equal to $30$ so that we get $b=16$. Substitute, we get numerator as
$$(x-3)(4x^2-4x-10)$$
From here, note that $4x^2-4x-10$ has no real roots, so we are good. Also we need $x=3$ to not be repeated root of denominator so we can say $a\neq 27$.
From the given limit, you can deduce that $$\lim_{x\to 3} \frac{4x^2-4x-10}{x^2+3x+9-a} = \frac{14}{27-a} = \frac{1}{2}$$
which gives $a = -1$.
No, not for all $a$.
You found a necessary condition, namely that $b=16$ in order to make the numerator $\to 0$ as $x\to 3$. The denominator will always $\to 0$ as $x\to 3$, but that does not mean that the limit of the quotient will be $\frac12$. But now that you have $b$, you can use l'Hopital to find the limit and see if you get a condition for $a$.