Chapter 2.3, Problem 13E

To determine

To evaluate: The limit of the function $\underset{t\to -3}{\lim }\frac{t^2-9}{2t^2+7t+3}$.

Answer to Problem 13E

The limit of the function is $\underset{\_}{\frac{6}{5}}$.

Explanation of Solution

Limit Laws:

Suppose that c is a constant and the limits $\underset{x\to a}{\lim }f\left(x\right)$ and $\underset{x\to a}{\lim }g\left(x\right)$ exist, then

Limit law 1:$\underset{x\to a}{\lim }\left[f\left(x\right)+g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }g\left(x\right)$

Limit law 2:$\underset{x\to a}{\lim }\left[f\left(x\right)-g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)-\underset{x\to a}{\lim }g\left(x\right)$

Limit law 7:$\underset{x\to a}{\lim }c=c$

Limit law 8:$\underset{x\to a}{\lim }x=a$

Limit law 9:$\underset{x\to a}{\lim }{x}^n=a^n$,  where n is a positive integer.

Direct substitution property:

“If f is a polynomial or a rational function and a is in the domain of f, then $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$”.

Fact 1:

If $f\left(x\right)=g\left(x\right)$ when $x\ne a$, then $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }g\left(x\right)$ , provided the limit exist.

Calculation:

The limit of the denominator is zero.

$\begin{array}{c}\underset{t\to -3}{\lim }\left(2t^2+7t+3\right)=\underset{t\to -3}{\lim }\left(2t^2\right)+\underset{t\to -3}{\lim }\left(7t\right)+\underset{t\to -3}{\lim }\left(3\right)\left(\text{by limit law 1}\right)\\ =2\underset{t\to -3}{\lim }\left(t^2\right)+7\underset{t\to -3}{\lim }\left(t\right)+\underset{t\to -3}{\lim }\left(3\right)\left(\text{by limit law 3}\right)\\ =2{\left(-3\right)}^2+7\left(-3\right)+3\left(\text{by limit law 9,8,7}\right)\\ =0\end{array}$

The quotient law cannot be used as the denominator of the function tends to zero when $t\to -3$.

Let $f\left(t\right)=\frac{t^2-9}{2t^2+7t+3}$ (1)

Simplify $f\left(t\right)$ by using elementary algebra.

Factorise the denominator $2t^2+7t+3$,

$\begin{array}{c}2t^2+7t+3=2t^2+6t+t+3\\ =2t\left(t+3\right)+1\left(t+3\right)\\ =\left(t+3\right)\left(2t+1\right)\end{array}$

Factorize the numerator $t^2-9$,

$t^2-9=\left(t+3\right)\left(t-3\right)$

Substitute $\left(t+3\right)\left(2t+1\right)$ for $2t^2+7t+3$ and $\left(t+3\right)\left(t-3\right)$ for $t^2-9$ in equation (1).

$f\left(x\right)=\frac{\left(t+3\right)\left(t-3\right)}{\left(t+3\right)\left(2t+1\right)}$

Cancel the common terms from both the numerator and the denominator, $f\left(t\right)=\frac{\left(t-3\right)}{\left(2t+1\right)}$.

Use fact 1, $f\left(t\right)=\frac{\left(t-3\right)}{\left(2t+1\right)}$ and $t\ne -3$, then $\underset{t\to -3}{\lim }\frac{t^2-9}{2t^2+7t+3}=\underset{t\to -3}{\lim }\frac{\left(t-3\right)}{\left(2t+1\right)}$.

Apply the direct substitution property on the limit function.

$\begin{array}{c}\underset{t\to -3}{\lim }\frac{\left(t-3\right)}{\left(2t+1\right)}=\frac{\left(-3-3\right)}{\left(2\left(-3\right)+1\right)}\\ =\frac{-6}{-6+1}\\ =\frac{-6}{-5}\\ =\frac{6}{5}\end{array}$

Thus, the limit of the function is $\underset{\_}{\frac{6}{5}}$.