Answer to Question #323762 in Calculus for Carmen

Question

given lim(as x approaches a) for ((x^2+bx+4b)/(x-a))= -6, find the values of a and b.

Answer 1

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.

$x^2+bx+4b=0$ if $x=a$ and due to factor theorem we can write:

$x^2+bx+4b=(x-a)(x-c)$ , where $c$ is some constant. To make limit equal to -6 we should choose $c=a+6$ .

$\displaystyle \lim_{\mathclap{x\to a}} {\frac{x^2+bx+4b}{x-a}}=\lim_{\mathclap{x\to a}} {\frac{(x-a)(x-a-6)}{x-a}}=$$\displaystyle\lim_{\mathclap{x\to a}} {(x-a-6)}=-6$

$x^2+bx+4b=(x-a)(x-a-6)$

$x^2+bx+4b=x^2-x(2a+6)+a^2+6a$

Equating coefficients, we get:

$b=-(2a+6)$

$4b=a^2+6a$

Substituting first equation into the second we will obtain:

$-4(2a+6)=a^2+6a$

$a^2+14a+24=0$

Using Vieta's formulas we can write

$a=-12$, $b=-(2a+6)=-(2\cdot(-12)+6)=18$

$a=-2$, $b=-(2\cdot(-2)+6)=-2$

Answer: $a=-12$ , $b=18$ or $a=-2$, $b=-2$ .