How to compute the limit,$\lim_{x\rightarrow 0}\frac{3x^2-3x\sin x}{x^2+x\cos1/x}$ How to compute the limit,
$$\lim_{x\rightarrow 0}\frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}$$
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 $a_n&gt;0$ 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 $\lim_{x\to x_0}f(x)=A\Leftrightarrow\forall \varepsilon&gt;0\exists\delta&gt;0\ s.t.\ 0&lt;|x-x_0|&lt;\delta\implies|f(x)-A|&lt;\varepsilon$
For any $\delta &gt; 0$ there exists an $0&lt;x&lt;\delta$ 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.