Question:

<p>\alpha = lim</p> u\to 0 eu -1 u = 1, because u = \sqrt tan x and v = \sqrt{x} both tend to 0, so the quotient is the derivative of eu at u = 0. For \beta, ln \beta = lim x\to 0 1 2 cot x \cdot ln(1 + sin x). Using ln(1 + sin x) ∼sin x, ln \beta = lim x\to 0 1 2 cot x \cdot sin x = lim x\to 0 1 2 cos x = 1<p><strong>2</strong>: </p> Hence, \beta = e1/2 = \sqrt{e.} Since the roots are 1 and \sqrt{e}, \alpha + \beta = 1 + \sqrt{e} = -b a, \alpha\beta = \sqrt{e} = - \sqrt{e} a . Thus a = -1, and so b = 1 + \sqrt{e.} Therefore, a + b = -1 + (1 + \sqrt{e}) = \sqrt{e.} Hence, 12 loge(a + b) = 12 loge(\sqrt{e}) = 12 \cdot 1 2 = 6. Shortcut / Fast View In limits of the form eu -ev u -v , think of the derivative of et.