# How can I determine the value of this limit? $\lim\limits_{t \to -\sqrt{8}} \frac{t + 2\sqrt{2}}{3t^2 - 24}.$

Evaluate the following limit:

$$\lim_{t \to -\sqrt{8}} \frac{t + 2\sqrt{2}}{3t^2 - 24}.$$

• I have tried L'Hopital rule but I have also tried factoring the denominator. I am not sure as to what answers are correct. Jun 6, 2014 at 15:11

$$\frac{t+2\sqrt 2}{3t^2-24}=\frac{t+\sqrt 8}{3(t+\sqrt 8)(t-\sqrt 8)}=\frac{1}{3(t-\sqrt 8)}$$Now apply the limit.

• Well understood, Thank you. Jun 6, 2014 at 15:34

Hint: $3t^2-24=3(t^2-8)=3(t+2\sqrt{2})(t-2\sqrt{2})$

Remark: Since you are familiar with L'Hospital's Rule, there is a more mechanical approach. First note that the Rule applies, top and bottom are approaching $0$. The derivative of the top is $1$, the derivative of the bottom is $6t$. And the limit of $\frac{1}{6t}$ as $t$ approaches $-2\sqrt{2}$ is clear.

• Well understood, thank you. Jun 6, 2014 at 15:21
• You are welcome. The L'Hospital's Rule method is easier in this case. But factorization remains an important technique. Jun 6, 2014 at 15:27

$$\lim_{t \to -2\sqrt{2}} \frac{t + 2\sqrt{2}}{3t^2 - 24} = \lim_{t \to -2\sqrt{2}} \frac{t + 2\sqrt{2}}{3(t^2 - 8)} =\lim_{t\to -2\sqrt 2} \frac{t + 2\sqrt 2}{3(t+ 2\sqrt 2)(t-2\sqrt 2)}$$

Cancel, and evaluate.

• The limit should be t approaching - sqrt 8 Jun 6, 2014 at 15:15
• It is: $$-\sqrt 8 = -\sqrt{4\cdot 2} = = -\sqrt 4\cdot \sqrt 2 = -2\sqrt 2$$ Jun 6, 2014 at 15:16
• Ok, sorry I've misunderstood that, so the answer should be -1/12 sqrt 2 Jun 6, 2014 at 15:20
• Yes, indeed, Kristi. Jun 6, 2014 at 15:22
• You're welcome, Kristi! Jun 6, 2014 at 15:38