Evaluate the following limit:
$$\lim_{t \to -\sqrt{8}} \frac{t + 2\sqrt{2}}{3t^2 - 24}.$$
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$\begingroup$ I have tried L'Hopital rule but I have also tried factoring the denominator. I am not sure as to what answers are correct. $\endgroup$– KristiJun 6, 2014 at 15:11
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3 Answers
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$$\frac{t+2\sqrt 2}{3t^2-24}=\frac{t+\sqrt 8}{3(t+\sqrt 8)(t-\sqrt 8)}=\frac{1}{3(t-\sqrt 8)}$$Now apply the limit.
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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.
answered Jun 6, 2014 at 15:10
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$\begingroup$ You are welcome. The L'Hospital's Rule method is easier in this case. But factorization remains an important technique. $\endgroup$ Jun 6, 2014 at 15:27
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$$\lim_{t \to -2\sqrt{2}} \frac{t + 2\sqrt{2}}{3t^2 - 24} = \lim_{t \to -2\sqrt{2}} \frac{t + 2\sqrt{2}}{3(t^2 - 8)} =\lim_{t\to -2\sqrt 2} \frac{t + 2\sqrt 2}{3(t+ 2\sqrt 2)(t-2\sqrt 2)}$$
Cancel, and evaluate.
