# How do you find #lim (3+u^(-1/2)+u^-1)/(2+4u^(-1/2))# as #u->0^+# using l'Hospital's Rule?

l"Hospital's Rule is neither needed nor helpful for this limit.

We no longer get an indeterminate form, so we cannot use (and have no need for) l'Hospital's Rule.

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To find the limit as ( u ) approaches ( 0^+ ) using L'Hôpital's Rule, we'll differentiate the numerator and the denominator with respect to ( u ) separately.

[ \lim_{u \to 0^+} \frac{3+u^{-1/2}+u^{-1}}{2+4u^{-1/2}} ]

First, differentiate the numerator and the denominator:

[ f(u) = 3 + u^{-1/2} + u^{-1} ] [ f'(u) = -\frac{1}{2}u^{-3/2} - u^{-2} ]

[ g(u) = 2 + 4u^{-1/2} ] [ g'(u) = -2u^{-3/2} ]

Now, apply L'Hôpital's Rule:

[ \lim_{u \to 0^+} \frac{f'(u)}{g'(u)} = \lim_{u \to 0^+} \frac{-\frac{1}{2}u^{-3/2} - u^{-2}}{-2u^{-3/2}} ]

[ = \lim_{u \to 0^+} \frac{-\frac{1}{2}u^{-3/2} - u^{-2}}{-2u^{-3/2}} ]

[ = \lim_{u \to 0^+} \frac{-\frac{1}{2} - u^{-1/2}}{-2} ]

[ = -\frac{1}{2} ]

So, the limit as ( u ) approaches ( 0^+ ) is ( -\frac{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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