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Extra resources for Calculus 1c-2, Examples of Elementary Functions
X ∈ ]1, +∞[), we get c = Arcsin 2x 1 + x2 + 2 Arctan x → 0 + 2 · π = π, 2 x → +∞. Since c is constant through the limit, we must have c = π, so Arcsin 2x 1 + x2 = π − 2 Arctan x, x ∈ ]1, +∞[. e. e. c = −π for x ∈ ] − ∞, −1[, and Arcsin 2x 1 + x2 = −π − 2 Arctan x, x ∈ ] − ∞, −1[. All things put together we see that ⎧ ⎨ π − 2 Arctan x, x ∈ ]1, +∞[, 2x 2 Arctan x, x ∈ ] − 1, 1[, Arcsin = ⎩ 1 + x2 −π − 2 Arctan x, x ∈ ] − ∞, −1[. Now, f (x) is continuous for every x ∈ R, and all the right hand sides are continuous in each their domains.
2) Apply (1) to prove that for every α > 0, ln x < 1 α x , α x > 0. 3) Prove for every β > 0 that ln x β < · x−β/2 , xβ 2 x > 0, and conclude that ln x →0 xβ for x → +∞, for every β > 0. The logarithm therefore increases signiﬁcantly slower that any power function of positive exponent. A. Investigate the growth of the logarithm and of the power functions. The procedure has been described in the text. D. Follow this procedure. 4 y 2 0 1 3 2 4 5 x –2 –4 Figure 17: The graphs of y = x and y = ln x, x > 0.
I. We must of course require that x ∈ [−1, 1]. If we put x = cos t, t ∈ [0, π], we see that this substitution is monotonous decreasing and that dx = − sin t dt and 1 − x2 = 1 − cos2 t = sin2 t = | sin t| = sin t for t ∈ [0, π]. Hence by insertion and an application of the fact that t = Arccos x, 1 − x2 dx = sin t · (− sin t) dt = − sin2 t dt 1 1 1 (1 − cos 2t) dt = − t + sin 2t 2 2 4 1 1 1 1 = − t + sin t · cos t = − t + cos t · 2 2 2 2 1 1 = − Arccos x + x 1 − x2 2 2 π 1 1 Arcsin x − + x 1 − x2 , = 2 4 2 = − 1 − cos2 t where the last result is either obtained from Arcsin x + Arccos x = π , 2 (the derivative is 0, and for x = 0 we get the value π2 ), or by using the substitution x = sin t, π π t ∈ − , , instead with dx = cos t dt.