1
$\begingroup$

Given $f(x) = [x + 1] (\sin(1/x))$, where[.] denotes greatest integer function ;

when $x\in (-1,0) \cup (0,1)$

$$f(x) = 0 , \text{ otherwise}$$

Question is to show f has discontinuity of second kind at $x=0$ and discontinuity of first kind at $x=1$

Attempt :At x=0 , LHL is zero , While RHL does not exist . I have seen this plugging values like 0.000000000001 and -0.00000000000001 .RHL does not exist because of lim as x goes to 0 sin(1/x) doesnt exist .But i also did this by putting $x = 0 + h$ and $x= 0 -h$ resp. Am i correct in drawing conclusions ? Thanks

$\endgroup$
1
$\begingroup$

HINT: $f(x)=\sin{1/x}$ for $x\in(0,1)$ (if $[x]$ means $\lfloor x \rfloor$).

$\endgroup$
  • $\begingroup$ thanks .What i have done ,is it correct ? $\endgroup$ – Sophie Clad Nov 23 '14 at 8:05
  • $\begingroup$ @SophieClad I suppose yes, if non-existence of limit in $0^+$ is obtained from properties of $\sin(1/x)$. (Your attempt gives no details). $\endgroup$ – Przemysław Scherwentke Nov 23 '14 at 8:11
  • $\begingroup$ Yes lim as x goes to zero of sin(1/x) DNE .i remember its graph very well .Thanks $\endgroup$ – Sophie Clad Nov 23 '14 at 8:30
0
$\begingroup$

You are on the right trace.

Note that the difinition of discontinuity of second kind is that one of LHL and RHl doesnot exist at least.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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