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.
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.