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Let f:(0,5) → ℝ be defined by f(x) = (√(5-x) - √5) / x. Show that f has a limit at x = 0. What is it?

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Final answer:

By rationalizing the expression for f(x), it is shown that the limit of f(x) as x approaches 0 is -1 / (2 * \sqrt{5}).

Step-by-step explanation:

The function f(x) = (\sqrt{5-x} - \sqrt{5}) / x is not defined at x = 0, but we can find its limit as x approaches 0. To do this, we use the conjugate of the numerator, which is \sqrt{5-x} + \sqrt{5}, to rationalize our expression.



Multiplying both the numerator and the denominator by the conjugate, we get:



(\sqrt{5-x} - \sqrt{5}) / x * (\sqrt{5-x} + \sqrt{5}) / (\sqrt{5-x} + \sqrt{5})



The product of the numerators will be:



(5-x - 5), which simplifies to -x.



Thus the original expression becomes:



-x / [x(\sqrt{5-x} + \sqrt{5})], and this simplifies to:



-1 / (\sqrt{5-x} + \sqrt{5}).



As x approaches 0, \sqrt{5-x} approaches \sqrt{5}, so the expression approaches:



-1 / (2\sqrt{5}) = -1 / (2 * \sqrt{5}).



Consequently, the limit of f(x) as x approaches 0 is -1 / (2 * \sqrt{5}).

User Hitesh Kansagara
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