Question

How do you find a value k such that the limit exists for the given function $\dfrac{{{x}^{2}}+8x+k}{x+2}$?

Answer 1

Hint: We start solving the problem by finding the value of x at which the given function is not defined. We then make use of the fact that the numerator must have a denominator as one of the factors to proceed through the problem. We then substitute $x=-2$ in ${{x}^{2}}+8x+k=0$ and make the necessary calculations to get the required value of k.

Complete step by step answer:
According to the problem, we are asked to find the value of k such that the limit exists for the given function $\dfrac{{{x}^{2}}+8x+k}{x+2}$.
We have given the function $\dfrac{{{x}^{2}}+8x+k}{x+2}$.
We can see that the function is not defined at $x=-2$, as the denominator becomes 0 for $x=-2$.
So, if the limit of the given function $\dfrac{{{x}^{2}}+8x+k}{x+2}$ exists, then the limits must exist at all values of the function. We can see that the limit at any value of the function can be found comfortably except at $x=-2$.
We know that the limit will exist at $x=-2$ if ${{x}^{2}}+8x+k$ has one of the factors as $x=-2$.
So, let us substitute $x=-2$ in ${{x}^{2}}+8x+k=0$ to find the value of k.
Now, we have ${{\left( -2 \right)}^{2}}+8\left( -2 \right)+k=0$.
$\Rightarrow 4-16+k=0$.
$\Rightarrow -12+k=0$.
$\Rightarrow k=12$.

$\therefore $ We have found the value of k such that the limit exists for the given function $\dfrac{{{x}^{2}}+8x+k}{x+2}$.

Note: Whenever we get this type of problems, we first try to find the value(s) of x at which the given function is not valid as we can have a possibility of discontinuity of the function at that value. Here the numerator and denominator of the given function are polynomials, we have opted for this method otherwise the method will vary. We can also find the value of k by assuming ${{x}^{2}}+8x+k=\left( x+2 \right)\left( x+\alpha \right)$ and expanding the multiplication to get the values of k and $\alpha $. Similarly, we can expect problems to find the value of $\alpha $ if limit exists for the function $\dfrac{\sin \left( \left( \alpha +1 \right)x \right)}{{{x}^{3}}}$.