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

Answer 1

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

We do get the indeterminate form #oo/oo#, so we can try l'Hospital's rile, but we'll get and continue to get negative powers of #u# in both the numerator and the denominator. Therefore, we will continue to get the indeterminate #oo/oo#
Multiply by #u/u#, to get
#(3+u^(-1/2)+u^-1)/(2+4u^(-1/2)) = (3u+u^(1/2)+1)/(2u+4u^(1/2))#

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

#lim_(xrarr0^+)(3u+u^(1/2)+1)/(2u+4u^(1/2)) = oo#
(The numerator is going to a positive limit and the denominator is going to #0# through positive value.)
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Answer 2

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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Answer from HIX Tutor

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.

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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