0

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?

user
  • 154,566
Joshua
  • 37
  • 2
  • Welcome to math stack exchange! As shown in the answer below, you can factor numerator and denominator and if the critical factor (here $t+3$) cancels out, you can determine the limit without L'Hopital – Peter Jun 09 '18 at 20:12
  • A tips for this sort of exercises/questions in school. Since you are asking to find the limit for $t \to -3$ and numerator vanishes at that $t$. It is very likely the denominator also vanishes. You should try to factor the denominator and then simplify the fraction. – achille hui Jun 09 '18 at 20:51
  • @Joshua Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Aug 03 '18 at 22:03

3 Answers3

6

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?

user
  • 154,566
2

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

Bernard
  • 175,478
1

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 $-6\epsilon$.

johnnyb
  • 3,509
  • Good catch. I meant that the whole term was infinitely close (which is what I did in the operation). Anyway, I’ll fix my language. – johnnyb Jun 10 '18 at 03:03