How do I find the limit without using L'Hopitals?

Question

If I have a function like

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

Is the only way to determine the limit to use L'Hopitals, or can it be done via algebra?

Answer 1

HINT

Note that for $t\neq -3$

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

then refer to Why are we allowed to cancel fractions in limits?

Answer 2

Yes: you simplify the fraction: $2t^2+7t+3=(t+3)(2t+1)$, so $$\frac{t^2-9}{2t^2+7t+3}=\frac{t-3}{2t+1}\to \frac 65\enspace\text{as } \:t\to-3.$$

Answer 3

For most limits, you can very easily just add in epsilon to the variable and solve directly (for right-handed limits - subtract to get left-handed limits). This is the method in non-standard analysis. Your teacher may or may not approve.

$$\lim_{t\to -3} \frac{t^2 - 9}{2t^2 + 7t + 3} = \frac{(t + \epsilon)^2 - 9}{2(t + \epsilon)^2 + 7(t + \epsilon) + 3} \\ = \frac{t^2 + 2t\epsilon + \epsilon^2 - 9}{2t^2 + 4t\epsilon + 2\epsilon^2 + 7t + 7\epsilon + 3} \\ = \frac{(-3)^2 + 2(-3)\epsilon + \epsilon^2 - 9}{2(-3)^2 + 4(-3)\epsilon + 2\epsilon^2 + 7(-3) + 7\epsilon + 3} \\ = \frac{-6\epsilon + \epsilon^2}{-5\epsilon + 2\epsilon^2 } \\ \approx \frac{-6\epsilon}{-5\epsilon} = \frac{6}{5} \\$$

The approximate at the end means "infinitely close to" since, I.e., $-6\epsilon + \epsilon^2$ is infinitely close to $-6\epsilon$.