Home » o-notation

Tag: o-notation

Prove that (1+x)c = 1 + cx + o(x) as x goes to 0

by RoRi

Prove that

    \[ \lim_{x \to 0} (1+x)^c = 1 + cx + o(x). \]

Using this fact compute

    \[ \lim_{x \to +\infty} \left( \sqrt{x^4 + x^2} - x^2 \right). \]

Proof. We use the definition and continuity of the exponential to evaluate the limit,

    \begin{align*} \lim_{x \to 0} (1+x)^c &= \lim_{x \to 0} e^{ c \log (1+x)} \\[9pt] &= \exp \left( \lim_{x \to 0} c \log (1+x) \right). \end{align*}

Then, we know (Examples on page 287 of Apostol, taking n = 1) that

    \[ \lim_{x \to 0} \log(1+x) = x - o(x). \]

Therefore,

    \begin{align*} \lim_{x \to 0} (1+x)^c &= \exp \left(\lim_{x \to 0} c \log(1+x) \right) \\[9pt] &= e^{ cx - o(x) }. \end{align*}

Finally we use the expansion, e^x = 1 + x + o(x), and Theorem 7.8 (page 288 of Apostol on the algebra of the o-symbols) to conclude,

    \begin{align*} \lim_{x \to 0}(1+x)^c &= e^{cx - o(x)} \\[9pt] &= 1 + (cx - o(x)) + o(cx-o(x)) \\[9pt] &= 1 + cx + o(x). \qquad \blacksquare \end{align*}

Now, to use this to evaluate the requested limit we make the substitution t = \frac{1}{x^2}. Then t \to 0 as x \to +\infty and we have

    \begin{align*} \lim_{x \to +\infty} \left( \sqrt{x^4+x^2} - x^2 \right) &= \lim_{t \to 0} \left( \sqrt{ \frac{1}{t^2} + \frac{1}{t} } - \frac{1}{t} \right) \\[9pt] &= \lim_{t \to 0} \left( \frac{1}{t} \sqrt{1 + t} - \frac{1}{t} \right) \\[9pt] &= \lim_{t \to 0} \frac{(1+t)^{\frac{1}{2}} - 1}{t} \\[9pt] &= \lim_{t \to 0} \frac{1 + \frac{1}{2} t + o(t) - 1}{t} \\[9pt] &= \lim_{t \to 0} \left( \frac{1}{2} + \frac{o(t)}{t} \right) \\[9pt] &= \frac{1}{2}. \end{align*}

Compute derivatives of a function given a limit equation that it satisfies

by RoRi

Consider a function f(x) that satisfies the limit relation

    \[ \lim_{x \to 0} \left( 1 + x + \frac{f(x)}{x} \right)^{\frac{1}{x}} = e^3, \]

and has a continuous third derivative everywhere. Determine the values

    \[ f(0), \quad f'(0), \quad f''(0), \quad \text{and} \quad \lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}}. \]

We do some simplification to the expression first.

    \begin{align*} &&\lim_{x \to 0} \left( 1 + x + \frac{f(x)}{x} \right)^{\frac{1}{x}} &= e^3 \\[9pt] \implies && \lim_{x \to 0} e^{\frac{1}{x} \log \left( 1 + x + \frac{f(x)}{x} \right)} &= e^3 \\[9pt] \implies && \exp \left( \lim_{x \to 0} \left( \frac{1}{x} \log \left( 1 +x + \frac{f(x)}{x} \right) \right) &= e^3 \\[9pt] \implies && \lim_{x \to 0} \left( \frac{1}{x} \log \left( 1 + x + \frac{f(x)}{x} \right) \right) &= 3. \end{align*}

Now, we apply the hint (that if \lim_{x \to 0} g(x) = A then g(x) = A + o(1)),

    \begin{align*} &&\frac{1}{x} \log \left( 1 + x + \frac{f(x)}{x} \right) &= 3 + o(1) \\[9pt] \implies && \log \left( 1 + x + \frac{f(x)}{x} \right) &= 3x + o(x) \\[9pt] \implies && 1 + x + \frac{f(x)}{x} &= e^{3x + o(x)} \\[9pt] \implies && x + x^2 + f(x) &= xe^{3x}e^{o(x)} \\[9pt] \implies && f(x) &= - x - x^2 + xe^{3x}e^{o(x)}. \end{align*}

So as x \to 0 we have

    \[ f(x) = -x - x^2 + x\left( 1 + 3x + o(x) \right) = 2x^2 + o(x^2). \]

But then since f has three derivatives at 0 we know its Taylor expansion at 0 is unique and is given by (Theorem 7.1, page 274 of Apostol)

    \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2} x^2 + o(x^2). \]

Hence, equating the coefficients of like powers of x we have

    \[ f(0) = 0, \quad f'(0) = 0, \quad f''(0) = 4. \]

Next, to compute the limit

    \[ \lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}} \]

we write

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

Then, using the Taylor expansion of \log(1+x) (page 287 of Apostol) we know as x \to 0 we have

    \[ \log (1+2x+o(x)) = 2x+o(x) + o(x) = 2x + o(x). \]

Therefore we have

    \begin{align*} \log \left( 1 + \frac{f(x)}{x} \right) = 2x + o(x) && \implies && 1 + \frac{f(x)}{x} &= e^{2x+o(x)} \\[9pt] && \implies && \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}} &= e^{2 + o(1)} \\[9pt] && \implies && \lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}} &= e^2. \end{align*}

Prove that 1 / (1+g(x)) = 1- g(x) + g2(x) + o(g2(x))

by RoRi
  1. Prove that

        \[ \frac{1}{1+g(x)} = 1 - g(x) + g^2(x) + o(g^2(x)) \quad \text{as} \quad x \to 0 \]

    where g(x) = o(1) as x \to 0.

  2. Using part (a) prove

        \[ \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + o(x^5) \quad \text{as} \quad x \to 0. \]

  1. Proof. We start with the algebraic identity

        \[ \frac{1}{1+u} = 1 - u + u^2 - \frac{u^3}{1+u}. \]

    Replacing u by g(x) we have,

        \[ \frac{1}{1+g(x)} = 1 - g(x) + g^2(x) - \frac{g^3(x)}{1+g(x)} = 1 - g(x) + g^2(x) - g^2(x) \frac{g(x)}{1+g(x)}. \]

    By the definition of o(1) we have

        \[ g(x) = o(1) \text{ as } x \to 0 \quad \implies \quad \lim_{x \to 0} g(x) = 0. \]

    Therefore,

        \[ \lim_{x \to 0} \frac{g(x)}{1+g(x)} = 0. \]

    Hence,

        \[ \frac{1}{1+g(x)} = 1 - g(x) + g^2(x) + o(g^2(x)). \qquad \blacksquare\]

  2. Proof. First, we use the expansion of \cos x,

        \[ \frac{1}{\cos x} = \frac{1}{1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^5)} = \frac{1}{1+g(x)} \]

    where g(x) = -\frac{x^2}{2} + \frac{x^4}{24} + o(x^5). By part (a) we then have as x \to 0,

        \begin{align*} \frac{1}{\cos x} &= 1 - g(x) + g^2(x) + o(g^2(x)) \\[9pt] &= 1 - \left( -\frac{x^2}{2} + \frac{x^4}{24} + o(x^5) \right) + \left( -\frac{x^2}{2} + \frac{x^4}{24} + o(x^5) \right)^2 + o \left( \left( -\frac{x^2}{2} + \frac{x^4}{24} \right)^2 \right) \\[9pt] &= 1 + \frac{x^2}{2} - \frac{x^4}{24} + \frac{x^4}{4} + o(x^4) \\[9pt] &= 1 + \frac{x^2}{2} + \frac{5x^4}{24} + o(x^4). \end{align*}

    Next, since \tan x = \frac{\sin x}{\cos x} we multiply this expansion for \frac{1}{\cos x} by the expansion (page 287 of Apostol) for \sin x as x \to 0,

        \begin{align*} \tan x &= \frac{1}{\cos x} \cdot \sin x \\[9pt] &= \left( 1 + \frac{x^2}{2} + \frac{5x^4}{24} + o(x^4) \right) \left( x - \frac{x^3}{6} + \frac{x^5}{120} + o(x^6) \right) \\[9pt] &= x - \frac{x^3}{3} + \frac{x^5}{120} + \frac{x^3}{2} - \frac{x^5}{12} + \frac{5x^5}{24} + o(x^5) \\[9pt] &= x + \frac{x^3}{6} + \frac{2x^5}{15} + o(x^5). \qquad \blacksquare \end{align*}

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

by RoRi

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,

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

    Hence,

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

    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.

    Next,

        \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.

Find the limit as x goes to 0 of (arcsin x / x)1/x2

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0} \left( \frac{\arcsin x}{x} \right)^{\frac{1}{x}}. \]

First, we rewrite the expression using the definition of the exponential:

    \[ \left( \frac{\arcsin x}{x} \right)^{\frac{1}{x}} = e^{\frac{1}{x^2} \log \left(\frac{\arcsin x}{x} \right)}. \]

Next, we need to get a series expansion for \arcsin x as x \to 0. The most straightforward way is to take the first few derivatives (we’ll only need the x^3 term).

    \begin{align*} f(x) &= \arcsin x & \implies && f(0) &= 0 \\[9pt] f'(x) &= \frac{1}{\sqrt{1-x^2}} & \implies && f'(0) &= 1 \\[9pt] f''(x) &= \frac{-x}{(1-x^2)^{\frac{3}{2}}} & \implies && f''(0) &= 0 \\[9pt] f'''(x) &= \frac{1+2x^2}{(1-x^2)^{\frac{5}{2}}} & \implies && f'''(0) &= 1. \end{align*}

Therefore, we have as x \to 0

    \[ \arcsin x = x + \frac{x^3}{6} + o(x^4). \]

Hence, as x \to 0

    \[ \frac{\arcsin x}{x} = 1 + \frac{x^2}{6} + o(x^3). \]

Since this is going to 1 as x \to 0 we may apply this exercise (Section 7.11, Exercise #4) to conclude

    \begin{align*} \log \left( \frac{\arcsin x}{x} \right) &= \left( \frac{\arcsin x}{x} - 1\right) - \frac{1}{2} \left( \frac{\arcsin x}{x} - 1 \right)^2 + o \left( \left( \frac{\arcsin x}{x} - 1 \right)^2 \right) \\[9pt] &= \left( \frac{x^2}{6} + o(x^3) \right) - \frac{1}{2} \left( \frac{x^2}{6} + o(x^3) \right)^2 + o \left( \left( \frac{x^2}{6} + o(x^3) \right)^2 \right) \\[9pt] &= \frac{x^2}{6} + o (x^3). \end{align*}

Now, we have the expansion as x \to 0

    \[ \frac{1}{x^2}\log \left( \frac{\arcsin x}{x} \right) = \frac{1}{6} + o (x). \]

Therefore,

    \begin{align*} \lim_{x \to 0} \left( \frac{\arcsin x}{x} \right)^{\frac{1}{x}} &= \lim_{x \to 0} e^{ \frac{1}{x^2} \log \left( \frac{\arcsin x}{x} \right)} \\[9pt] &= e^{\frac{1}{6}} \lim_{x \to 0} e^{o(x)} \\ &= e^{\frac{1}{6}}. \end{align*}