# How do you find a value k such that the limit exists given (x^2+8x+k)/(x+2)?

Aug 18, 2017

If we define:
$L \left(\alpha\right) = {\lim}_{x \rightarrow \alpha} \frac{{x}^{2} + 8 x + k}{x + 2}$

If $\alpha = - 2$ the limit exists provided $k = 12 \implies L \left(- 2\right) = 4$

For any other value of alpha, the limit always exist, $k$ is arbitrary, and

$L \left(\alpha\right) = \frac{{\alpha}^{2} + 8 \alpha + k}{\alpha + 2}$

#### Explanation:

The question is incomplete as the value of $x$ is not specified for the limit.

Let us consider the function:

$L \left(\alpha\right) = {\lim}_{x \rightarrow \alpha} \frac{{x}^{2} + 8 x + k}{x + 2}$

The limit function is not defined when the denominator is zero, ie when $x + 2 = 0 \implies x = - 2$

So let us consider the case $\alpha = - 2$, in which case we examine:

$L \left(- 2\right) = {\lim}_{x \rightarrow - 2} \frac{{x}^{2} + 8 x + k}{x + 2}$

This limit will exist iff the numerator also has a factor of $\left(x + 2\right)$, along with another factor (x+beta), say, This would require:

$\left(x + \beta\right) \left(x + 2\right) \equiv {x}^{2} + 8 x + k$
$\implies {x}^{2} + \left(2 + \beta\right) x + 2 \beta = {x}^{2} + 8 x + k$

If we equate coefficients then we get:

${x}^{1} : 2 + \beta = 8 \implies \beta = 6$
${x}^{0} : 2 \beta = k \implies k = 12$

Hence, we have a factor that will cancel with $\beta = 6 , k = 12$ givings

$L \left(- 2\right) = {\lim}_{x \rightarrow - 2} \frac{{x}^{2} + 8 x + 12}{x + 2}$
$\text{ } = {\lim}_{x \rightarrow - 2} \frac{\left(x + 6\right) \left(x + 2\right)}{x + 2}$
$\text{ } = {\lim}_{x \rightarrow - 2} \left(x + 6\right)$
$\text{ } = 4$

For any other value of $\alpha$, the limit will always exist, ie $k$ is arbitrary, and

$L \left(\alpha\right) = {\lim}_{x \rightarrow \alpha} \frac{{x}^{2} + 8 x + k}{x + 2}$
$\text{ } = \frac{{\alpha}^{2} + 8 \alpha + k}{\alpha + 2}$