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Virty [35]
3 years ago
9

Lim t^4 - 6 / 2t^2 - 3t + 7

Mathematics
1 answer:
Harman [31]3 years ago
6 0

I think you meant to say

\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}

(as opposed to <em>x</em> approaching 2)

Since both the numerator and denominator are continuous at <em>t</em> = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}

Because these expressions are continuous at <em>t</em> = 2, we can compute the limits by evaluating the limands directly at 2:

\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}

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