Question

Is there a number\(a\)such that

\(\mathop {\lim }\limits_{x \to - 2} \frac{{3{x^2} + ax + a + 3}}{{{x^2} + x - 2}}\)

exists? If so, find the value of \(a\) and the value of the limit.

Short answer

The value of \(a\) is \(15\) and value of limit is \( - 1\).

Step-by-step solution

Analyze the given function

It can be noted that as x tends to\( - 2\), the value of the denominator approaches 0 as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to - 2} \,{x^2} + x - 2 &= {\left( { - 2} \right)^2} + \left( { - 2} \right) - 2\\ &= 4 - 2 - 2\\ &= 0\end{aligned}\)

Thus, for limit to exist at \(x = - 2\), the numerator must also approach 0 as x tends to \( - 2\). For this, the following equation must be true:

\(\mathop {\lim }\limits_{x \to - 2} \left( {3{x^2} + ax + a + 3} \right) = 0\)

Solve the resulting equation

Apply the limit by plugging\(x = - 2\)in the expression and solve for\(a\), as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to - 2} \left( {3{x^2} + ax + a + 3} \right) = 0\\3{\left( { - 2} \right)^2} + a\left( { - 2} \right) + a + 3 = 0\\12 - 2a + a + 3 = 0\\a= 15\end{aligned}\).

Find the limit

Finally, solve the given limit by plugging\(a = 15\), as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to - 2} \frac{{3{x^2} + 15x + 18}}{{{x^2} + x - 2}} &= \mathop {\lim }\limits_{x \to - 2} \frac{{3\left( {x + 2} \right)\left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}}\\ &= \mathop {\lim }\limits_{x \to - 2} \frac{{3\left( {x + 3} \right)}}{{\left( {x - 1} \right)}}\\ &= \frac{{3\left( { - 2 + 3} \right)}}{{ - 2 - 1}}\\ &= \frac{3}{{ - 3}}\\ &= - 1\end{aligned}\)

Thus, the value of a limit is 15.