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

Try getting common denominators in the left product.

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To find the limit using L'Hospital's Rule, we first notice that as ( t ) approaches ( 0^+ ), both ( \frac{1}{t} ) and ( \sqrt{t} ) approach infinity. By applying L'Hospital's Rule, we differentiate the numerator and the denominator separately until we can evaluate the limit.

First, we differentiate the numerator and the denominator: [ \lim_{t \to 0^+} \frac{1/t + 1/\sqrt{t}}{\sqrt{t+1} - 1} ] [ = \lim_{t \to 0^+} \frac{-1/t^2 - 1/(2t\sqrt{t})}{1/(2\sqrt{t+1})} ]

Then, we simplify: [ = \lim_{t \to 0^+} \frac{-2\sqrt{t+1} - \sqrt{t}}{t^{3/2}} ]

As ( t ) approaches ( 0^+ ), both ( \sqrt{t+1} ) and ( \sqrt{t} ) approach 1. Therefore, the limit is: [ = -2 \times 1 = -2 ]

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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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