Question

Evaluate the limit if it exists.

$\underset{t\to -3}{\lim }\frac{t^2-9}{2t^2+7t+3}$

Short answer

The value of limit of the given function is$\frac{6}{5}.$

Step-by-step solution

Step 1. Given information.

The limit of function is given here,

$\underset{t\to -3}{\lim }\frac{t^2-9}{2t^2+7t+3}$

Step 2. Formula used.

The given function is simplified by applying the formula.

Formula:$\left(a^2-b^2\right)=(a+b)(a-b)$

Step 3. Simplification.

Given, $\underset{t\to -3}{\lim }\frac{t^2-9}{2t^2+7t+3}$

$\begin{array}{c}\underset{t\to -3}{\lim }\frac{t^2-9}{2t^2+7t+3}=\frac{{\left(t\right)}^2-{\left(3\right)}^2}{2t^2+(6+1)t+3}\\ =\frac{(t+3)(t-3)}{2t^2+6t+t+3}\\ =\frac{(t+3)(t-3)}{2t(t+3)+1(t+3)}\\ =\frac{(t+3)(t-3)}{(2t+1)(t+3)}\\ =\frac{t-3}{2t+1}\\ \end{array}$

Now, we apply the value:

$\begin{array}{c}=\frac{-3-3}{2\times -3+1}\\ =\frac{-6}{-6+1}\\ =\frac{-6}{-5}\\ =\frac{6}{5}\end{array}$

Therefore, our final answer is$\frac{6}{5}.$