# Calculate limit for, $\lim\limits_{x\to 0}\frac{1-cos(x^6)}{x^{12}}$, but in there have suprize.

Let's think about this function, $\quad \to f(x)=\dfrac{x^2-1}{x-1}$,

$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=0/0$ ,

First Solution :

$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=\lim\limits_{x\to 1}\dfrac{x+1}{1}$

$=\lim\limits_{x\to 1}\dfrac{x+1}{1}=\dfrac{1+1}{1}=2$

And Second Solution with l'hopital:

$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}\overbrace{\longrightarrow}^{l'hopital}\lim\limits_{x\to 1}\dfrac{2x}{1}=2$

But...

Let's we take this function ,$f(x)=\dfrac{1-cos(x^6)}{x^{12}}$

$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=0/0$,

For this funciton couldn't simplification,I have just , graph of function and l'hôpital,

1-L'hôpital;

$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=\lim\limits_{x\to 0}\dfrac{6.x^5.sin(x^6)}{12.x^{11}}=\underbrace{\lim\limits_{x\to 0}\dfrac{sin(x^6)}{x^6}}_1.\dfrac{1}{2}=\dfrac{1}{2}$

2-Graph of function;

Graph telling me ,$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=0$

l'hôpital $\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=1/2$

What we do now?

And link of graph , if you can't see clearly,https://www.desmos.com/calculator/jfz4kslm4w

• Ok, so what is the question, again? – DonAntonio Jul 15 '16 at 17:32
• The limit is $1/2$. Graphing programs often misbehave. In this case it may be roundoff error in the arithmetic engine. The limit is clearly the same as the limit of $(1-\cos u)/u^2$. The graphing program will probably behave better on this. – André Nicolas Jul 15 '16 at 17:32
• General question. Would the limit become easier to calculate if you perform a substitution $x^6=t$ and let $t$ approach zero? Just a thought – imranfat Jul 15 '16 at 17:47
• @Photoneaterman I love it:) – imranfat Jul 15 '16 at 17:54
• @imranfat Thank you, glad to see you around. – DonAntonio Jul 15 '16 at 18:12

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

This is the usual procedure to deal with a function in any computer language. It yields the $\ul{right\ plot}$ of the function !!!.

1. The machine-precision $\mathtt{pm}$ ( which is language dependent ) I'm using right now is $\mathtt{pm} = 2.84217 \times 10^{-14}$.
2. I took the variable $\mathtt{tol} = 2.98428 \times 10^{-14}$ which is $5\ \%$ above $\mathtt{pm}$ in order to prevent rounding errors.
3. $${1 - \cos\pars{x^{6}} \over x^{12}} \approx \half - {x^{12} \over 24} + {x^{24} \over 720} - {x^{36} \over 40320}\,,\qquad \verts{x} \gtrsim 0$$I 'cut' this expression whenever $\ds{{x^{36} \over 40320} < \mathtt{tol}\ \imp\ \verts{x} < 0.565239}$. Moreover, I avoid the calculation of $\ds{1 \over x^{12}}$ whenever $\ds{{1 \over x^{12}} \leq \mathtt{tol}\ \imp\ \verts{x} \geq \mathtt{tol}^{-1/12} = 13.3998}$.
4. The function I plot is given by $$\left\lbrace\begin{array}{lcl} \ds{\half} & \mbox{if} & \ds{x = 0} \\[2mm] \ds{\half - {x^{12} \over 24}\pars{1 - {x^{12} \over 30}}} & \mbox{if} & \ds{0 < \verts{x} \leq 0.565239} \\[2mm] \ds{1 - \cos\pars{x^{6}} \over x^{12}} & \mbox{if} & \ds{0.565239 < \verts{x} < 13.3998} \\[2mm] \ds{0} & \mbox{if} & \ds{\verts{x} \geq 13.3998} \\&& \end{array}\right.$$

Note that the polynomial is evaluated with the Horner Rule and $\ul{never}$ as $\ds{\half - {x^{12} \over 24} + {x^{24} \over 720}}$ !!!. Since $\ds{6 = 2 \times 3}$ and $\ds{12 = 2 \times 2 \times 3}$, $x^{6}$ and $\ds{x^{12}}$ are evaluated in a few steps.

For example, in $\mathtt{C++}$:

double x,x3,x6,x12;
x3 = x*x*x;
x6 = x3*x3;
x12 = x6*x6;


The $\ul{right\ plot}$:

• Isn't this solution assuming the limit (even more, the Taylor expansion to the third non-zero term, cf. 3. and 4.) of the function, while the question is about finding this limit? – Clement C. Jul 17 '16 at 14:22
• @ClementC. You are right and the 'trivial' limit was already found in another answers. I $\underline{want\ to\ concentrate}$ in the "surprise" factor the OP claimed. User $\mathrm{@}$André Nicolas already pointed out, in a comment, that the 'arithmetic' evaluation of the function was not right enough. Thanks for your remark. – Felix Marin Jul 17 '16 at 19:03
• I see. Thanks for the clarification! – Clement C. Jul 17 '16 at 19:07

Hint: \begin{align} \frac{1-\cos\left(x^6\right)}{x^{12}} &=\frac{1-\cos\left(x^6\right)}{\sin^2\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\ &=\frac{1-\cos\left(x^6\right)}{1-\cos^2\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\ &=\frac1{1+\cos\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\ \end{align}

• Very nice way to show the limit without l'Hospital. +1 – DonAntonio Jul 15 '16 at 17:33

The weird oscillations and behavior you see near 0 in the graph are due to rounding/precision errors by the computer. Near $x=0$, both $1-\cos(x^6)$ and $x^{12}$ are very very small numbers that machines will have trouble handling. Asking a machine to divide one small number by another small number is just asking for trouble.

To see what the graph should look like near $x=0$, simply take the first two terms of the Taylor expansion about 0: $$\frac{1-\cos(x^6)}{x^{12}} \approx \frac{1}{2} - \frac{x^{12}}{24}$$

For anything remotely close to 0, it will just look like a flat function at $1/2$.

In short, the limit from L'hopital is correct and the graph you have generated is not accurate near $x = 0$ due to precision limits of computers.

• @Nukeguy. This is why it is so important to teach the students the limitations of this great technology. My students are "hooked up" to the Ti83/84, which I consider a very nice tool. Now let the TI come up with the derivative at the left hand side of zero for $y=arctan(1/x)$ through the calc mode. With the limit definition you can verify that the curve leaves the y-axis with a slope of $-1$, but when you type in the TI 0.00000001 it will completely mess up. I have a whole bucket of problems like that for the class to make them realize that we still need to do the rigor. – imranfat Jul 15 '16 at 17:53

The short version is that graphing technology can misbehave for very small $x$-values. That is what is happening here and your graph is just incorrect (A fault of the software, not the way you used it).

For small values of $x$, $\cos(x^6)$ is $1$ minus an extremely small value. The software is probably using floating point decimals, so it records an initial digit in the tenths place, and only keeps track of decimals out so far.

Meanwhile $x^{12}$ in the denominator keeps track of the same number of significant digits, but it starts tracking them far past the decimal.

When it carries out the subtraction in the numerator, it is not reaching a result that has the appropriate relative precision. It's still essentially starting its significant figure count with the tenths place, and so it's not keeping track of digits out to the order of where $x^{12}$ has its first nonzero digit.

Imagine if a computer could only track three significant digits. Let's look at \begin{align} \frac{1-\cos(0.001^6)}{0.001^{12}}&\cong\frac{1.00-1.00}{0.001^{12}} \end{align}

Since this is the best it can do for evaluating $\cos(0.001^6)$. It carries on with the arithmetic and gets $0$. In your situation the tolerance is probably more than three significant digits, but it's the same idea.

• Thank you a lot , for interesting . Good . – Azer BABA fanlari Jul 17 '16 at 6:51

Two basic limits everyone should know from high school are $$\lim_{u\to 0}\frac{1-\cos u}u=0\qquad\text{and}\qquad\lim_{u\to 0}\frac{1-\cos u}{u^2}=\frac12$$ whereby a simple substitution leads to the result.

Hints for the proof:

Rewrite the fractions as $$\frac{(1-\cos u)(1+\cos u)}{u^{(2)}(1+\cos u)}=\frac{\sin^2u}{u^{(2)}(1+\cos u)}.$$

• Thank you for interesting . – Azer BABA fanlari Jul 17 '16 at 6:51

Setting $x^6=2y$ and using $\cos2y=1-2\sin^2y$ to find $$\lim_{y\to0^+}\dfrac{1-\cos2y}{(2y)^2}=\dfrac24\left(\lim_{y\to0^+}\dfrac{\sin y}y\right)^2=?$$

• $\boxed{\text{Thank you a lot , for interesting.}}$ – Azer BABA fanlari Jul 17 '16 at 6:52