TITLE: The continuity of a function at a point $ x = 0$
QUESTION [5 upvotes]: Choose correct options , more than one may be correct .
(1)

$$ \textrm{The function  defined by } \begin{cases} f(x)=\cos\left(\dfrac{1}{x}\right) & x\neq 0\\ f(0)=0 & \\ \end{cases} \qquad \textrm{is continuous at x=0.} $$

(2)

$$ \textrm{The function  defined by } \begin{cases} f(x)=\sin x\sin\left(\dfrac{1}{x}\right) & x\neq 0\\ f(0)=0 & \\ \end{cases} \qquad \textrm{is continuous at x=0.} $$

(3) 

$$ \textrm{The function  defined by } \begin{cases} f(x)=x+\sin\left(\dfrac{1}{x}\right) & x\neq 0\\ f(0)=1 & \\ \end{cases} \qquad \textrm{is continuous at x=0.} $$

(4)  

$$ \textrm{The function  defined by } \begin{cases} f(x)=x\sin\left(\dfrac{1}{x}\right) & x\neq 0\\ f(0)=1 & \\ \end{cases} \qquad \textrm{is continuous at x=0.} $$

here is graph of each function  but i don't know how to plot it exactly via wolfram
for (1) : Graph of function (1) via Wolfram

for(2) : Graph of function (2) via wolfram

for(3) : Graph of function (3) via wolfram

for(4) : Graph of function (4) via wolfram

I think the correct answer is (2) Indeed :
It is well know, that 
$$\left|\sin{x}\right|\leq 1 \quad \forall x \in \mathbb{R}$$
Because of this it holds 
$$\left| \sin x \sin\dfrac{1}{x}\right|\leq \left|\sin x\right|\cdot 1 = \left|\sin x\right|$$
Thus
$$\lim_{x\rightarrow 0} f(x) = 0 = f(0) $$
which means that $f$ is continous at $x=0$ 

Would you please show me why others options aren't correct ?
How can i plot those functions in wolfram

Thanks and Regards.

REPLY [2 votes]: Quick:
$$\lim_{x\to\pm\infty}\cos x\,\;or\;\,\sin x\;\;\;\;\;\text{don't exist}$$
and this already takes off options (1) and (3).
In (4) we have a function that goes to zero ($\;x\;$) times a bounded function $\;\left(\sin\frac1x\right)\;$ and thus the limit is zero, yet it doesn't equal the defined value of the function there thus (4) not true.