How do you find #lim (1+5/sqrtu)/(2+1/sqrtu)# as #u->0^+# using l'Hospital's Rule?

Answer 1
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Answer 2

# lim_(u rarr 0^+) (1+5/sqrt(u))/(2+1/sqrt(u)) = 5 #

We do not need to apply L'Hôpital's as this is a trivial limit to evaluate:

# lim_(u rarr 0^+) (1+5/sqrt(u))/(2+1/sqrt(u)) = lim_(u rarr 0^+) (1+5/sqrt(u))/(2+1/sqrt(u))*sqrt(u)/sqrt(u) # # " " = lim_(u rarr 0^+) (sqrt(u)+5)/(2sqrt(u)+1) # # " " = 5/1 #
Note that the above limit is only valid for #u rarr 0^+# as:
# u rarr 0^+ => u > 0 => sqrt(u) in RR #

Whereas:

# u rarr 0^- => u < 0 => sqrt(u) in CC #
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Answer 3

To find ( \lim_{u \to 0^+} \frac{1 + \frac{5}{\sqrt{u}}}{2 + \frac{1}{\sqrt{u}}} ) using L'Hôpital's Rule:

  1. Notice that as ( u \to 0^+ ), both the numerator and denominator approach ( \infty ).
  2. Apply L'Hôpital's Rule by taking the derivative of the numerator and denominator separately with respect to ( u ).
  3. Differentiate the numerator and denominator and simplify.
  4. Evaluate the limit of the resulting expression as ( u \to 0^+ ).
  5. The resulting value is the limit of the original expression.

By applying L'Hôpital's Rule, you can find the limit of the given expression as ( u ) approaches ( 0^+ ).

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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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