DIAMMIBENVERMk1

2022-07-07

How to find the value of a and b from this limit problem with or without L'Hopital's formula?
Consider the limit:
$\underset{x\to 4}{lim}\frac{{x}^{2}+ax+b}{x-4}=14$
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:
$\underset{x\to 4}{lim}\frac{2x+a}{1}=14$
Then, we can substitute the limit of x to the equation such that:
$2\left(4\right)+a=8+a=14$
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

Expert

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
$\underset{x\to 4}{lim}\frac{\left(x-\alpha \right)\left(x-\beta \right)}{x-4}=14$
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
$\underset{x\to 4}{lim}\left(x-\beta \right)=14$
which can be easily solved to give $\beta =-10$
Now equating the coefficients of
${x}^{2}+ax+b=\left(x-\alpha \right)\left(x-\beta \right)=\left(x-4\right)\left(x+10\right)$
gives the required values for a and b.

sweetymoeyz

Expert

You need that limit to have the $0/0$ indeterminate form. So
${4}^{2}+4a+b=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}b=-16-4a.$
But you found out that $a=6$. So $b=-40$

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