Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 Graphe of function (1)

for(2) : Graph of function (2) via wolfram enter image description here

for(3) : Graph of function (3) via wolfram enter image description here

for(4) : Graph of function (4) via wolfram enter image description here

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
Thanks would you please take look to second question –  Adam Apr 5 at 20:16
    
How can you plot those in WA? I've no idea...but aren't those diagrams in your question from WA?? –  DonAntonio Apr 5 at 20:18
    
yes it's from WA but i can 't add the second term f(0)=0 –  Adam Apr 5 at 20:20
    
The limit exists for (4), so the limit as $x \to 0$ exists for $(\sin x/x) (x \sin(1/x))$. Recheck option (2). –  nayrb Apr 5 at 20:54
1  
And if you will, please try not to be so emphatic. I'm not trying to be mean nor am I trying to be a dunce. –  nayrb Apr 6 at 15:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.