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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.

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2
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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.

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  • $\begingroup$ Thanks would you please take look to second question $\endgroup$ – Adam Apr 5 '14 at 20:16
  • $\begingroup$ How can you plot those in WA? I've no idea...but aren't those diagrams in your question from WA?? $\endgroup$ – DonAntonio Apr 5 '14 at 20:18
  • $\begingroup$ yes it's from WA but i can 't add the second term f(0)=0 $\endgroup$ – Adam Apr 5 '14 at 20:20
  • $\begingroup$ The limit exists for (4), so the limit as $x \to 0$ exists for $(\sin x/x) (x \sin(1/x))$. Recheck option (2). $\endgroup$ – abnry Apr 5 '14 at 20:54
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    $\begingroup$ 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. $\endgroup$ – abnry Apr 6 '14 at 15:48

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