Home » o-notation » Page 2

Tag: o-notation

Find the limit as x goes to 0 of (x + e2x)1/x

Evaluate the limit.

    \[ \lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}}. \]


From the definition of the exponential we have

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

So, first we use the expansion of e^x as x \to 0 (page 287 of Apostol) to write

    \[ e^{2x} = 1 + 2x + o(x) \qquad \text{as} \quad x \to 0. \]

Therefore, as x \to 0 we have

    \[ \log (x + e^{2x}) = \log (x + 1  + 2x + o(x)) = \log (1 + 3x + o(x)). \]

Now, since 3x + o(x) \to 0 as x \to 0 we can use the expansion (again, page 287) of \log (1+x) as x \to 0 to write

    \[ \log (1 + 3x + o(x)) = 3x+o(x) + o(3x+o(x)) = 3x + o(x) \qquad \text{as} \quad x \to 0. \]

Therefore, as x \to 0 we have

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

So, getting back to the expression we started with,

    \[ \lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = \lim_{x \to 0} e^{3 + \frac{o(x)}{x}} = e^3 \lim_{x \to 0} e^{\frac{o(x)}{x}}. \]

But, as in the previous exercise (Section 7.11, Exercise #23) we know \lim_{x \to 0} e^{\frac{o(x)}{x}} = 1. Hence,

    \[ \lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = e^3. \]

Find the limit as x goes to 1 of x1 / (1-x)

Evaluate the limit.

    \[ \lim_{x \to 1} x^{\frac{1}{1-x}}. \]


First, we write

    \[ x^{\frac{1}{1-x}} = e^{\frac{1}{1-x} \cdot \log x}. \]

From this exercise (Section 7.11, Exercise #4) we know that as x \to we have

    \[ \log x = (x-1) + o((x-1)) \]

Therefore, as x \to 1,

    \[ \frac{\log x}{1-x} = -1 + \frac{o(x-1)}{x-1}. \]

So, we then have

    \begin{align*}  \lim_{x \to 1} x^{\frac{1}{1-x}} &= \lim_{x \to 1}e^{\frac{\log x}{1-x}} \\[9pt]  &= \lim_{x \to 1} e^{-1 + \frac{o(x-1)}{x-1}} \\[9pt]   &= \lim_{x \to 1} e^{-1} \cdot e^{\frac{o(x-1)}{x-1}} \\[9pt]  &= \frac{1}{e} \lim_{x \to 1} e^{\frac{o(x-1)}{x-1}}. \end{align*}

(Here we could say that since the exponential is a continuous function we can bring the limit inside and so this becomes e^0 = 1. I’m not sure we know we can pass limits through continuous functions like that, so we continue on with expanding the exponential as in previous exercises.)
Since \frac{o(x-1)}{x-1} \to 0 as x \to 1 we take the expansion of e^x as x \to 0,

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

Therefore,

    \[ \lim_{x \to 1} x^{\frac{1}{1-x}} = \frac{1}{e} \lim_{x \to 1} e^{\frac{o(x-1)}{x-1}} = \frac{1}{e}. \]

Find the limit as x goes to 0 of (ax – asin x) / x3

Evaluate the limit.

    \[ \lim_{x \to 0} \frac{a^x - a^{\sin x}}{x^3}. \]


First, we want to get expansions for a^x and a^{\sin x} as x \to 0. For a^x we write a^x = e^{x \log a} and use the expansion (page 287 of Apostol) of e^x. This gives us

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

Next, for a^{\sin x}, again we write a^{\sin x} = e^{\sin x \log a} and then use the expansion for e^x we have

    \[ a^{\sin x} = e^{\sin x \log a} = 1 + (\log a \sin x) + \frac{(\log a)^2}{2} (\sin x)^2 + \frac{(\log a)^3}{6} (\sin x)^3 + o((\sin x)^3). \]

Now, we need use the expansion for \sin x (again, page 287 of Apostol)

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

and substitute this into our expansion of a^{\sin x},

    \begin{align*}  a^{\sin x} &= 1 + (\log a \sin x) + \frac{(\log a)^2}{2} (\sin x)^2 + \frac{(\log a)^3}{6} (\sin x)^3 + o((\sin x)^3) \\[9pt]  &= 1 + (\log a) \left( x - \frac{x^3}{6} + o(x^4) \right) + \frac{(\log a)^2}{2} \left( x - \frac{x^3}{6} + o(x^4) \right)^2 \\  & \qquad + \frac{(\log a)^3}{6} \left( x - \frac{x^3}{6} + o(x^4) \right)^3 + o\left( \left( x - \frac{x^3}{6} + o(x^4) \right)^3 \right) \\[9pt]  &= 1 + (\log a) x + \frac{(\log a)^2}{2} x^2 + \left( -\frac{\log a}{6} + \frac{(\log a)^3}{6} \right) x^3 + o(x^3). \end{align*}

(Again, this is the really nice part of little o-notation. We had lots of terms in powers of x greater than 3, but they all get absorbed into o(x^3), so we don’t actually have to multiply out and figure out what they all were. We only need to figure out the terms for the powers of x up to 3. Of course, the 3 could be any number depending on the situation; we chose 3 in this case because we know that’s what we will want in the limit we are trying to evaluate.)

So, now we have expansions for a^x and a^{\sin x} (in which most of the terms cancel when we subtract) and we can evaluate the limit.

    \begin{align*}  \lim_{x \to 0} \frac{a^x - a^{\sin x}}{x^3} &= \lim_{x \to 0} \frac{\frac{\log a}{6} x^3 + o(x^3)}{x^3} \\[9pt]  &= \lim_{x \to 0} \left(\frac{\log a}{6} + \frac{o(x^3)}{x^3}\right)\\[9pt]  &= \frac{\log a}{6}. \end{align*}