# How do you find the limit of #(t^2-4)/(t^3-8)# as t approaches 2?

is possible to use L'Hôpital's rule

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An algebraic solution:

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To find the limit of (t^2-4)/(t^3-8) as t approaches 2, we can substitute 2 into the expression and simplify. Plugging in t=2, we get (2^2-4)/(2^3-8), which simplifies to (4-4)/(8-8). This further simplifies to 0/0. Since we have an indeterminate form, we can apply L'Hôpital's rule by taking the derivative of the numerator and denominator separately. The derivative of t^2-4 is 2t, and the derivative of t^3-8 is 3t^2. Evaluating these derivatives at t=2, we get 2(2)=4 and 3(2^2)=12. Now we can find the limit of the derivatives, which is 4/12 = 1/3. Therefore, the limit of (t^2-4)/(t^3-8) as t approaches 2 is 1/3.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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