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

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To find the limit of the expression (1 - t/(t - 1)) / (1 - sqrt(t)/(t - 1)) as t approaches infinity using L'Hospital's Rule:

- Rewrite the expression as lim t->∞ (1 - t/(t - 1)) / (1 - sqrt(t)/(t - 1)).
- Since both the numerator and denominator approach infinity as t approaches infinity, apply L'Hospital's Rule by taking the derivative of the numerator and the derivative of the denominator separately.
- Take the derivative of the numerator: d/dt (1 - t/(t - 1)) = d/dt (1 - 1 + 1/(t - 1)) = d/dt (1/(t - 1)).
- Take the derivative of the denominator: d/dt (1 - sqrt(t)/(t - 1)) = d/dt (1 - t^(-1/2)/(t - 1)) = d/dt (1 - (t^(-1/2))/(t - 1)).
- Apply L'Hospital's Rule again if necessary until you reach a determinable limit or a form that can be simplified easily.
- Evaluate the limit of the derivatives as t approaches infinity.
- The result is the limit of the original expression as t approaches infinity.

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To find ( \lim_{t \to \infty} \frac{1 - \frac{t}{t-1}}{1 - \frac{\sqrt{t}}{t-1}} ) using L'Hôpital's Rule, we apply the rule to the indeterminate form ( \frac{0}{0} ).

First, we differentiate the numerator and denominator separately with respect to ( t ):

[ \lim_{t \to \infty} \frac{d}{dt} \left(1 - \frac{t}{t-1}\right) \div \frac{d}{dt} \left(1 - \frac{\sqrt{t}}{t-1}\right) ]

[ = \lim_{t \to \infty} \frac{d}{dt} \left(1 - \frac{t}{t-1}\right) \div \frac{d}{dt} \left(1 - \frac{\sqrt{t}}{t-1}\right) ]

[ = \lim_{t \to \infty} \frac{0 - (t-1)(-1) - (-t)(t-1)^{-2}}{0 - (t-1)^{-\frac{3}{2}} - (-\frac{1}{2}t^{-\frac{1}{2}})(t-1)^{-1}} ]

[ = \lim_{t \to \infty} \frac{(t-1) - (t-1)^{-2}t}{(t-1)^{-\frac{3}{2}} + \frac{1}{2}t^{-\frac{1}{2}}(t-1)^{-1}} ]

[ = \lim_{t \to \infty} \frac{(t-1)^2 - t}{(t-1)^{-\frac{3}{2}} + \frac{1}{2}(t-1)^{-\frac{3}{2}}} ]

[ = \lim_{t \to \infty} \frac{t^2 - 2t + 1 - t}{(t-1)^{-\frac{3}{2}} + \frac{1}{2}(t-1)^{-\frac{3}{2}}} ]

[ = \lim_{t \to \infty} \frac{t^2 - 3t + 1}{(t-1)^{-\frac{3}{2}} + \frac{1}{2}(t-1)^{-\frac{3}{2}}} ]

Now, as ( t \to \infty ), the denominator ( (t-1)^{-\frac{3}{2}} + \frac{1}{2}(t-1)^{-\frac{3}{2}} ) approaches ( 0 ), and the numerator ( t^2 - 3t + 1 ) grows without bound. Therefore, the limit is of the form ( \frac{\infty}{0} ), which is an indeterminate form. So, we can apply L'Hôpital's Rule again.

Differentiate the numerator and denominator separately:

[ = \lim_{t \to \infty} \frac{d}{dt} (t^2 - 3t + 1) \div \frac{d}{dt} \left((t-1)^{-\frac{3}{2}} + \frac{1}{2}(t-1)^{-\frac{3}{2}}\right) ]

[ = \lim_{t \to \infty} \frac{2t - 3}{-\frac{3}{2}(t-1)^{-\frac{5}{2}} - \frac{3}{4}(t-1)^{-\frac{5}{2}}} ]

[ = \lim_{t \to \infty} \frac{2t - 3}{-\frac{3}{2}(t-1)^{-\frac{5}{2}} - \frac{3}{4}(t-1)^{-\frac{5}{2}}} ]

[ = \lim_{t \to \infty} \frac{2t - 3}{-\frac{9}{4}(t-1)^{-\frac{5}{2}}} ]

[ = \lim_{t \to \infty} -\frac{8}{9} \cdot (t-1)^{\frac{5}{2}} ]

Now, as ( t \to \infty ), ( (t-1)^{\frac{5}{2}} ) also goes to ( \infty ), but with a negative sign. So, the limit of ( -\frac{8}{9} \cdot (t-1)^{\frac{5}{2}} ) is ( -\infty ).

Therefore, the limit ( \lim_{t \to \infty} \frac{1 - \frac{t}{t-1}}{1 - \frac{\sqrt{t}}{t-1}} ) is ( -\infty ).

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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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