How to compute the limit,$\lim_{x\rightarrow 0}\frac{3x^2-3x\sin x}{x^2+x\cos1/x}$

Question

How to compute the limit, $$\lim_{x\rightarrow 0}\frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}$$

Answer 1

Rewrite $$\lim_{x\to 0}\frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}=\lim_{x\to 0}\frac{3x^2-3x\sin x}{x(x+\cos\frac{1}{x})},$$ we will deduce the nature of $x+\cos\frac1x$ near $0$.
There are monotone decreasing sequence $a_n>0$ with $\lim_{n\to\infty}a_n=0$ such that $\cos\frac 1{a_n}=(-1)^n$.
Let $a_n\ll 1$, we have $a_{n+1}+\cos \frac 1{a_{n+1}}$ have a different sign from $a_{n}+\cos \frac 1{a_{n}}$. Since $(x+\cos\frac{1}{x})$ is continuous, there is a zero point between $a_{n+1}$ and $a_n$.
Hence there are infinite undefined points near $0$. Therefore, the limit does not exist.
Notice that this answer is only available with definition $\lim_{x\to x_0}f(x)=A\Leftrightarrow\forall \varepsilon>0\exists\delta>0\ s.t.\ 0<|x-x_0|<\delta\implies|f(x)-A|<\varepsilon$

Answer 2

This is a tricky problem. The first issue as pointed out in other answers is that the denominator vanishes in every deleted neighborhood of $0$ so that in every deleted neighborhood of $0$ there are points where the function is not defined. The usual definition of limit requires that the function be defined in some deleted neighborhood of the point under consideration and hence based on this definition the limit does not exist. In fact one should say that it does not make sense to talk of the limit of this function based on this definition as the prerequisites of the definition are not fulfilled.

However there is a relaxed definition of limit which considers only points of domain of definition of the function:

Let $D $ be a non-empty subset of $\mathbb {R} $ and let $a\in\mathbb{R} $ be an accumulation point of $D$. Let $f:D\to \mathbb{R}$ be a function. A number $L$ is said to be the limit of $f$ at $a$ if for any given $\epsilon >0$ there exists a corresponding $\delta >0$ such that $|f(x) - L|<\epsilon $ whenever $x\in(a-\delta, a+\delta) \cap D, x\neq a$.

The limit in question does not exist even under this relaxed definition of limit but it is difficult to show that.

Let $g(x) =x^4-x$ and note that $g(x) <0$ if $0<x<1$ and $g(x) \to 0$ as $x\to 0^{+}$. Further as $x\to 0^{+}$ the function $\cos(1/x)$ oscillates between $1$ and $-1$ and therefore by intermediate value theorem $\cos(1/x)=g(x)$ or $x+\cos (1/x)=x^4$ for infinitely many values of $x$ as $x\to 0^{+}$.

The given function in question can be written as $$3\cdot\frac {x-\sin x} {x+\cos (1/x)}$$ and the numerator behaves like $x^3/2$ as $x\to 0^{+}$. As discussed in previous paragraph there is a sequence of positive values of $x$ tending to $0$ for which denominator is equal to $x^4$ and thus if $x\to 0^{+}$ through such a sequence of values then the given function tends to $\infty$. On the other hand using similar strategy we can find a sequence of positive values of $x$ such that denominator equals $x^3$ so that the fraction tends to $1/2$. Thus the given function oscillates infinitely as $x\to 0^{+}$ and the desired limit does not exist.

The technique used above can be applied to prove that given any number $L$ there is a sequence $x_n$ such that $x_n\to 0$ and $f(x_n) \to L$ as $n\to\infty $. One can also have $L=\pm\infty$. The cases $L=1/2,L=\infty$ were discussed above.

Answer 3

As noticed the limit doesn't exist indeed we can consider the sequence as $x_n \to 0$ such that

$$\cos\frac{1}{x_n}=2x_n \implies \frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}=\frac{3x_n^2-3x_n\sin x_n}{3x_n^2}=1-\frac{\sin x_n}{x_n} \to 1-1=0$$

and the sequence $x_n \to 0$ such that

$$\cos\frac{1}{x_n}=-x_n+x_n^3 \implies \frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}=3\frac{x_n^2-x_n\sin x_n}{x_n^4}=3\frac{x_n-\sin x_n}{x_n^3}\to \frac12$$

indeed as $t \to 0$ we have that $\frac{t-\sin t}{t^3} \to \frac16$ wich can be proved by l'Hopital, Taylor or by the method shown here: Are all limits solvable without L'Hôpital Rule or Series Expansion.

For the issue already discussed here in detail by Paramanand Singh about the not satisfactory way to show that the limit doesn't exist because not defined at "infinitely many points", for a useful discussion refer also to the related

• What is $\lim_{x \to 0}\frac{\sin(\frac 1x)}{\sin (\frac 1 x)}$ ? Does it exist?

Answer 4

For any $\delta > 0$ there exists an $0<x<\delta$ such that $x - \cos \frac {1}{x} = 0$

As $x-\cos \frac 1x$ is continuous, the are points close to these zeros such that $ x-\cos \frac 1x $ is arbitrarily close to zero, and $\frac {3x^2 - 3x\sin x}{x(x-\cos \frac 1x)}$ is arbitrarily large.