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Prove or disprove given statements for functions such that f(x) = o(g(x))

Let f and g be functions, both differentiable in a neighborhood of 0, with g(x) > 0 and such that

    \[ f(x) = o(g(x)) \qquad \text{as} \qquad x \to 0. \]

Prove or disprove each the following statements.

  1. \displaystyle{ \int_0^x f(t) \, dt = o \left( \int_0^x g(t) \, dt \right)} as x \to 0.
  2. f'(x) = o(g'(x)) as x \to 0.

  1. True.
    Proof. Since f(x) = o(g(x)) as x \to 0 we know by the definition of o(g(x)) that

        \[ \lim_{x \to 0} \frac{f(x)}{g(x)} = 0. \]

    Thus, for every \varepsilon > 0 there exists a \delta > 0 such that

        \[ |x| < \delta \quad \implies \quad \left| \frac{f(x)}{g(x)} \right| < \varepsilon. \]

    So, for |x| < \delta we have

        \begin{align*}  \left| \frac{\int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} \right| &\leq \frac{\int_0^x |f(t)| \, dt}{\left| \int_0^x g(t) \, dt \right|} \\[9pt]  &< \frac{\varepsilon \int_0^x g(t) \, dt }{\left| \int_0^x g(t) \, dt \right|} \\[9pt]  &= \varepsilon. \end{align*}

    The final line follows since g > 0 by hypothesis. Therefore,

        \[ |x| < \delta \quad \implies \quad \left| \frac{\int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} \right| < \varepsilon. \]


        \[ \lim_{x \to 0} \frac{ \int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} = 0. \]

    By definition, we then have

        \[ \int_0^x f(t) \, dt = o \left( \int_0^x g(t) \, dt \right). \qquad \blacksquare\]

  2. False.
    Consider f(x) = x^2 \sin \left( \frac{1}{x} \right) for x \neq 0 and f(x) = 0 for x = 0. Then, for x \neq 0,

        \[ f'(x) = 2x \sin \left( \frac{1}{x} \right) - \cos \left( \frac{1}{x} \right). \]

    For x = 0 we have f'(0) = 0.


        \begin{align*}  f(x) &= x^2 \sin \left( \frac{1}{x} \right) \\[9pt]  &= x^2 \left( \frac{\sin \left( \frac{1}{x} \right)}{\frac{1}{x}} \right) \left( \frac{1}{x} \right) \\[9pt]  &= x \left( \frac{\sin \left( \frac{1}{x} \right)}{\frac{1}{x}} \right). \end{align*}

    Since \lim_{x \to 0} \frac{\sin (1/x)}{1/x} = 1 we have f(x) = o(x) as x \to 0. However, f'(x) \neq o(1) since

        \[ \lim_{x \to 0} \left( 2 \sin \left( \frac{1}{x}\right) - \cos \left( \frac{1}{x} \right) \right) \]

    does not exist.


  1. hteica says:

    I think part (b) could use a more simpler counterexample. For example, let $f(x)=x$ and $g(x)=1$ defined for all real numbers. Then $f$ and $g$ both have derivatives in some interval containing 0, and clearly $g$ is always positive, also $f(x)=o(g(x))$ as $x\rightarrow 0$. But we can see that $1=f^{\prime}(x)=o(g^{\prime}(x))=o(0)$ is not true, since o(0) is undefined.

    • Artem says:

      This counterexample is actually quite interesting :) I wonder if it is legit to conclude the case from the non-existence of o(0). If you have a question somewhere on stack, feel free to share for everyone :)

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