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Properties of functions.

  1. Nov 19, 2007 #1
    Among the following function,
    -one is continuous everywhere,but not differentiable at 0
    -one is differentiable,but derivative is discontinuous at 0
    -one is differentiable and has continuous derivative

    1)[tex]f(x) = \left\{\begin{array}{cc}\ln (1 + x^3)\sin\frac {1}{x},x > 0 \\

    2)[tex]g(x) = \left\{\begin{array}{cc}\ln^2(1 + x)\sin\frac {1}{x},x > 0 \\

    3)[tex]h(x) = \left\{\begin{array}{cc}\ln (1 + \frac {\sin x}{2}),x > 0 \\


    1)[tex]\lim_{x\rightarrow 0}\frac {\ln (1 + x^3)\sin\frac {1}{x}}{x} = \lim_{x\rightarrow 0}\frac {(x^3 + 0{x^6})\sin\frac {1}{x}}{x} = 0[/tex]


    [tex]f'(x) = \frac {3x^2\sin\frac {1}{x}}{1 + x^3} - \frac {\cos\frac {1}{x}\ln (1 + x^3)}{x^2}[/tex]

    2)[tex]\lim_{x\rightarrow 0}\frac {(x^2 + o(x^4))\sin\frac {1}{x}}{x} = 0[/tex] differentiable

    [tex]g'(x) = \frac {2\ln (1 + x)\sin\frac {1}{x}}{1 + x} - \frac {\cos\frac {1}{x}\ln^2 (1 + x)}{x^2}[/tex]

    3)[tex]\lim_{x\rightarrow 0}\frac {(\frac {\sin x}{2} + 0(\sin^2x))\sin\frac {1}{x}}{x} = \lim_{x\rightarrow 0}\sin\frac {1}{x}[/tex] not differentiable

    I couldn't find which function f'(x) or g'(x) is not differentiable at 0,which one continuous.
  2. jcsd
  3. Nov 20, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    This may not be the most analytical method, but try graphing these functions, it becomes easier to see.
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