How to find the value of $a$ and $b$ from this limit problem with or without L'Hopital's formula?

Question

Consider the limit:

$$\lim_{x\to4} \frac{x^2+ax+b}{x-4} =14$$

Question: How can I find the values of $a$ and $b$?

Attempt:

My first thought is, we need to use L'Hopital's rule to make sure that the denominator isn't zero:

Applying L'Hopital's rule, and we get:

$$\lim_{x\to4}\frac{ 2x+a}1 = 14$$

Then, we can substitute the limit of $x$ to the equation such that:

$$2(4)+a = 8+a =14$$

and we get that the value of $a$ is $6$.

But, how can I find the value of $b$? It seems that after applying the L'Hopital's formula the value of $b$ disappears.

Also, is there a way to solve this problem without L'Hopital's rule?

Thanks

Answer 1

You need that limit to have the $0/0$ indeterminate form. So $$4^2 + 4a + b = 0 \implies b = -16 - 4a.$$

But you found out that $a = 6 $. So $b = -40 $.

Answer 2

Observe the denominator tends to $\;0\;$ when $\;x\to 4\;$ , but since the limit is finite it must be that also the numerator apporaches zero, so $\;x=4\;$ has to be a zero of $\; x^2+6x+b\;$ . End the argument now.

Answer 3

To solve this without L'Hopital, observe that since the top of the fraction in the limit is a quadratic polynomial, we can rewrite the limit as $$\lim_{x \to 4}\frac{(x-\alpha)(x-\beta)}{x-4} = 14$$ where $\alpha, \beta \in \mathbb C$ are the roots of the polynomial $x^2 + ax + b$.

In order for the limit to exist, as $x\to 4$, since the bottom of the fraction tends to $0$, we must have $x^2 + ax +b \to 0$, so in particular, at least one of $\alpha, \beta$ must be $4$. Say $\alpha$ is $4$ (swapping the order doesn't change anything).

Then the limit becomes $$\lim_{x \to 4}(x-\beta) = 14$$which can be easily solved to give $\beta = -10$.

Now equating the coefficients of $$x^2 + ax + b = (x-\alpha)(x-\beta) = (x-4)(x+10)$$ gives the required values for $a$ and $b$.