# How do you find #lim (t+1/t)((4-t)^(3/2)-8)# as #t->0# using l'Hospital's Rule?

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To find the limit ( \lim_{t \to 0} (t + \frac{1}{t})((4 - t)^{\frac{3}{2}} - 8) ) using L'Hôpital's Rule, we first rewrite the expression to an indeterminate form ( \frac{0}{0} ), then take the derivative of the numerator and the derivative of the denominator separately.

- Rewrite the expression:

[ \lim_{t \to 0} (t + \frac{1}{t})((4 - t)^{\frac{3}{2}} - 8) ]

- Expand the expression:

[ = \lim_{t \to 0} \frac{(t + \frac{1}{t})(\sqrt{4 - t} - 2\sqrt{2})(\sqrt{4 - t} + 2\sqrt{2})}{(t + \frac{1}{t})} ]

[ = \lim_{t \to 0} (\sqrt{4 - t} - 2\sqrt{2})(\sqrt{4 - t} + 2\sqrt{2}) ]

- Apply L'Hôpital's Rule to the indeterminate form:

[ = \lim_{t \to 0} \frac{d}{dt}((\sqrt{4 - t} - 2\sqrt{2})(\sqrt{4 - t} + 2\sqrt{2})) ]

[ = \lim_{t \to 0} \frac{d}{dt}(4 - t - 8) ]

[ = \lim_{t \to 0} (-1) ]

[ = -1 ]

So, ( \lim_{t \to 0} (t + \frac{1}{t})((4 - t)^{\frac{3}{2}} - 8) = -1 ).

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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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