Tag: Taylor Polynomials

Find the limit as x goes to 0 of (cos (sin x) – cos x) / x⁴

by RoRi

December 22, 2015

Evaluate the limit.

$\lim_{x \to 0} \frac{\cos (\sin x) - \sin x}{x^4}.$

We know (page 287 of Apostol) that the expansions for $\sin x$ and $\cos x$ as $x \to 0$ are given by

$\begin{align*} \sin x &= x - \frac{x^3}{6} + o(x^4) \\ \cos x &= 1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^5). \end{align*}$

Therefore, we have the following expansion for $\cos (\sin x)$ as $x \to 0$ ,

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

So, now we can take the limit,

$\begin{align*} \lim_{x \to 0} \frac{\cos (\sin x) - \sin x}{x^4} &= \lim_{x \to 0} \frac{1 - \frac{1}{2}x^2 + \frac{5}{24}x^4 + o(x^4) - 1 + \frac{1}{2}x^2 - \frac{1}{24}x^4 + o(x^5)}{x^4} \\[9pt] &= \lim_{x \to 0} \frac{\frac{1}{6} x^4 + o(x^4)}{x^4} \\[9pt] &= \lim_{x \to 0} \left( \frac{1}{6} + \frac{o(x^4)}{x^4} \right) \\[9pt] &= \frac{1}{6}. \end{align*}$

Find the limit as x goes to 0 of (a^x – a^{sin x}) / x³

by RoRi

December 22, 2015

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.

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

Evaluate the limit as x goes to 0 of a given function

by RoRi

December 22, 2015

Evaluate the limit.

$\lim_{x \to 0} \frac{3 \tan (4x) - 12 \tan x}{3 \sin (4x) - 12 \sin x}.$

We use the expansions for $\tan x$ (given in Example 1 on page 288 of Apostol) and $\sin x$ (on page 287 of Apostol):

$\begin{align*} \tan x &= x + \frac{1}{3}x^3 + o(x^3) \\ \tan (4x) &= 4x + \frac{64}{3}x^3 + o(x^3) \\ \sin x &= x - \frac{1}{6}x^3 + o(x^3) \\ \sin (4x) &= 4x - \frac{32}{3}x^3 + o(x^3). \end{align*}$

Therefore, we have

$\begin{align*} \lim_{x \to 0} \frac{3 \tan (4x) - 12 \tan x}{3 \sin (4x) - 12 \sin x} &= \lim_{x \to 0} \frac{3 \left( 4x + \frac{64}{3} x^3 + o(x^3) \right) - 12 \left( x + \frac{1}{3} x^3 + o(x^3) \right)}{ 3 \left( 4x - \frac{32}{3} x^3 + o(x^3) \right) - 12 \left( x - \frac{1}{6}x^3 + o(x^3) \right)} \\[10pt] &= \lim_{x \to 0} \frac{12x + 64x^3 + o(x^3) - 12x - 4x^3 + o(x^3)}{12x - 32x^3 + o(x^3) - 12x + 2x^3 + o(x^3)} \\[10pt] &= \lim_{x \to 0} \frac{60x^3 + o(x^3)}{-30x^3 + o(x^3)} \\[10pt] &= \lim_{x \to 0} \frac{60 + \frac{o(x^3)}{x^3}}{-30 + \frac{o(x^3)}{x^3}} \\[10pt] &= -2. \end{align*}$

Find the limit as x goes to 0 of (cosh x – cos x) / x²

by RoRi

December 22, 2015

Evaluate the limit.

$\lim_{x \to 0} \frac{\cosh x - \cos x}{x^2}.$

We use the definition of $\cosh x$ in terms of the exponential:

$\cosh x = \frac{e^x + e^{-x}}{2},$

and the expansions (page 287 of Apostol) of $e^x$ and $\cos x$ as $x \to 0$ :

$\begin{align*} e^x &= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3) \\ e^{-x} &= 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + o(x^3) \\ \cos x &= 1 - \frac{x^2}{2} + o(x^3). \end{align*}$

Putting these together we evaluate the limit:

$\begin{align*} \lim_{x \to 0} \frac{\cosh x - \cos x}{x^2} &= \lim_{x \to 0} \frac{e^x + e^{-x} - 2\cos x}{2x^2} \\[9pt] &= \lim_{x \to 0} \frac{1 + x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3) + 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + o(x^3) - 2 \cos x}{2x^2} \\[9pt] &= \lim_{x \to 0} \frac{2 + x^2 + o(x^3) - 2\left(1 - \frac{x^2}{2} + o(x^3) \right)}{2x^2} \\[9pt] &= \lim_{x \to 0} \frac{2x^2 + o(x^3)}{2x^2} \\[9pt] &= \lim_{x \to 0} \left( 1 + \frac{o(x^3)}{x^2} \right) \\[9pt] &= 1. \end{align*}$

Find the limit as x goes to 1 of given function

by RoRi

December 22, 2015

Evaluate the limit.

$\lim_{x \to 1} \frac{\sin \left( \frac{\pi}{2x} \right) \cdot \log x}{(x^3+5)(x-1)}.$

Using the expansion (from this exercise, Section 7.11, Exercise #4)

$\log x = (x-1) + o(x-1)$

we compute the limit as follows:

$\begin{align*} \lim_{x \to 1} \frac{\sin \left( \frac{\pi}{2x} \right) \cdot \log x}{(x^3 + 5)(x-1)} &= \lim_{x \to 1} \left( \frac{\sin \left( \frac{\pi}{2x} \right)}{x^3+5} \right) \left(\frac{\log x}{x-1} \right) \\[9pt] &= \frac{1}{6} \lim_{x \to 1} \frac{\log x}{x-1} \\[9pt] &= \frac{1}{6} \lim_{x \to 1} \frac{(x-1) + o(x-1)}{x-1} \\[9pt] &= \frac{1}{6}. \end{align*}$

Find the limit as x goes to (π / 2) of cos x / (x – (π / 2))

by RoRi

December 21, 2015

Evaluate the limit.

$\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{x - \frac{1}{2} \pi}.$

Since $\cos x = -\sin \left( x - \frac{\pi}{2} \right)$ we use the expansion (page 287) for $\sin x$ ,

$\sin \left( x - \frac{\pi}{2} \right) = \left( x - \frac{\pi}{2} \right) + o \left( \left( x - \frac{\pi}{2} \right)^2 \right).$

Therefore,

$\begin{align*} \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{x - \frac{\pi}{2}} &= \lim_{x \to \frac{\pi}{2}} \frac{ -\left(x - \frac{\pi}{2}\right) + o\left( \left( x - \frac{\pi}{2} \right)^2 \right)}{x - \frac{\pi}{2}} \\[9pt] &= -1. \end{align*}$

Find the limit as x goes to 0 of the given expression

by RoRi

December 21, 2015

Evaluate the limit.

$\lim_{x \to 0} \frac{\log(1+x) - x}{1 - \cos x}.$

We use the expansions on page 287 for $\log(1+x)$ and $\cos x$ ,

$\begin{align*} \log (1+x) &= x - \frac{x^2}{2} + \frac{x^3}{3} + o(x^3) \\ \cos x &= 1 - \frac{x^2}{2} + o (x^3). \end{align*}$

Then we compute the limit,

$\begin{align*} \lim_{x \to 0} \frac{ \log(1+x) - x}{1 - \cos x} &= \lim_{x \to 0} \frac{-\frac{x^2}{2} + \frac{x^3}{3}+ o(x^3)}{\frac{x^2}{2} + o(x^3)} \\[9pt] &= \lim_{x \to 0} \frac{ -\frac{1}{2} + \frac{x}{3} + \frac{o(x^3)}{x^2}}{\frac{1}{2} + \frac{o(x^3)}{x^2}} \\[9pt] &= -1. \end{align*}$

Evaluate the limit as x goes to 0 of (1 – cos (x²)) / (x² sin (x²))

by RoRi

December 21, 2015

Evaluate the limit.

$\lim_{x \to 0} \frac{1 - \cos (x^2)}{x^2 \sin(x^2)}.$

We use the expansions (p. 287) for $\cos x$ and $\sin x$ as $x \to 0$ to write,

$\begin{align*} \cos (x^2) &= 1 - \frac{x^4}{2!} + o(x^5) \\ \sin (x^2) &= x^2 + o(x^4) \\ \end{align*}$

Therefore, we compute the limit as

$\begin{align*} \lim_{x \to 0} \frac{1 - \cos (x^2)}{x^2 (\sin (x^2))} &= \lim_{x \to 0} \frac{1 - 1 + \frac{1}{2}x^4 + o(x^5)}{x^2(x^2+o(x^4))} \\[9pt] &= \lim_{x \to 0} \frac{\frac{1}{2}x^4 + o(x^5)}{x^4 + o(x^6)} \\[9pt] &= \lim_{x \to 0} \frac{\frac{1}{2} + \frac{o(x^5)}{x^4}}{1 + \frac{o(x^6)}{x^4}} \\[9pt] &= \frac{1}{2}. \end{align*}$

Find the limit as x goes to 1 of log x / (x² + x – 2)

by RoRi

December 21, 2015

Evaluate the limit.

$\lim_{x \to 1} \frac{\log x}{x^2 + x -2}.$

From this exercise (Section 7.11, Exercise #4) we know

$\log x = (x-1) - \frac{1}{2}(x-1)^2 + o((x-1)^2).$

Therefore,

$\begin{align*} \lim_{x \to 1} \frac{\log x}{x^2+x-2} &= \lim_{x \to 1} \frac{(x-1) - \frac{1}{2}(x-1)^2 + o((x-1)^2)}{x^2+x-2} \\[9pt] &= \lim_{x \to 1} \left( \frac{1}{x+2} + \frac{(x-1)}{2(x+2)} + \frac{o(x-1)^2}{x-1}\cdot \frac{1}{x+2} \right) \\[9pt] &= \frac{1}{3} + 0 + 0 \cdot \frac{1}{3} \\[9pt] &= \frac{1}{3}. \end{align*}$

Find the limit as x goes to 0 of (a^x – 1) / (b^x – 1)

by RoRi

December 21, 2015

Evaluate the limit for $b \neq 1$ .

$\lim_{x \to 0} \frac{a^x - 1}{b^x - 1}.$

First we write $a^x = e^{x \log a}$ and $b^x = e^{x \log b}$ . Then we use the expansion of $e^x$ (p. 287), to obtain expansions for $a^x$ and $b^x$ ,

$a^x = e^{x \log a} = 1 + (x \log a) + o (x), \qquad b^x = e^{x \log b} = 1 + (x \log b) + o(x).$

Therefore, we have

$\begin{align*} \lim_{x \to 0} \frac{a^x - 1}{b^x - 1} &= \lim_{x \to 0} \frac{1 + x \log a + o(x) - 1}{1 + x \log b + o(x) - 1} \\[9pt] &= \lim_{x \to 0} \frac{x \log a + o(x)}{x \log b + o(x)} \\[9pt] &= \lim_{x \to 0} \frac{\log a + \frac{o(x)}{x}}{\log b + \frac{o(x)}{x}} \\[9pt] &= \frac{\log a}{\log b}. \end{align*}$