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The point $t=0$, for the function $f(t) = (\sin t)/t$ is

a) a discontinuity point

b) a maximum

c) a minimum.

I marked a) as the option because even though right hand side and left hand limit exists for this function which is $1$ but $f(0)$ is not defined.

However answer is b) .

Can someone please tell me why is a) incorrect?

Paolo Intuito
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Ramesh
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2 Answers2

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This function is not defined at 0, so it can't be discontinuous there (a point of continuity or discontinuity must belong to the domain).

If you define $f(0)$ as 1, since $1=\lim_{x\to 0}f(x)$, then $f$ has a maximum at 0, because $|x|>|\sin x|$ for all nonzero $x$ (this is well known inequality, discussed many times).

Without having $f(0)$ defined, $f$ has no maximum (then it takes values arbitrarily close to 1, but without 1).

tong_nor
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  • What if domain would have been [-180 , +180] ? Then will f(t) be discontinuous ? – Ramesh Nov 21 '17 at 15:44
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    0 is in this interval, if you put 0 into the function you get $\frac{0}{0}$ which is undefined, so you have to define $f(0)$ somehow to have your function well defined on this interval. – tong_nor Nov 21 '17 at 15:48
  • I am sorry but I didn't quite understand this. You said we should define a domain right and check if all points are continuos within that domain ? So 0 lies inside the domain and it is discontinuous right ? I am not getting this. – Ramesh Nov 21 '17 at 15:51
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    the given function has $(-\infty,0)\cup(0,+\infty)$ as its domain, since $f(0)$ has no sense. If you want to define $f$ on an interval containing $0$, then you first have to define $f(0)$ somehow. Since the limit exists, there is natural definition of $f(0)$ as the limit. – tong_nor Nov 21 '17 at 15:58
  • Thanks for the reply. So if it's 0/0 then it's undefined. If it would have been 1/0 then it's discontinuous right? – Ramesh Nov 21 '17 at 16:08
  • The function is discontinuous at x=0 is it not? And isn't that called a removable discontinuity? – Jose M Serra Nov 21 '17 at 16:35
  • A point of continuity or discontinuity first must belong to the domain. Until you don't define $f(0)$, it's impossible to say if it is continuous at 0 or not. – tong_nor Nov 21 '17 at 16:41
  • Here something to help http://math.mit.edu/~jspeck/18.01_Fall%202014/Supplementary%20notes/01c.pdf – Jose M Serra Nov 21 '17 at 16:45
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I disagree with the answer and using a Calculus textbook James Stewart Essential Calculus on page 46 it states

"If f is defined near a (in other words, f is defined on an open interval containing a except perhaps at a) we say that f is discontinuous at a (or f had a discontinuity at a) if f is not continuous at a."

In your case every t before 0 is defined and every t after 0 therefore the function is discontinuous at t=0.

Graph of function: https://www.desmos.com/calculator/hbcw8ht2fg

Jose M Serra
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    I think everyone has different opinion on this. For eg see this answer https://math.stackexchange.com/questions/1454781/is-this-function-fx-x-1-3-continuous-when-x-varies-from-1-to-1 – Ramesh Nov 21 '17 at 17:52
  • @RajeshR Yeah the function is continuous for all values except at x=0 because its a removable discontinuity, and the other case is a vertical asymptote so your connection is wrong. Have you even seen the graph of the function? – Jose M Serra Nov 21 '17 at 21:24
  • If we expand using the series of sinx, then x gets cancelled and we get maximum value of 1 on y axis. – Ramesh Nov 22 '17 at 06:57
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    @RajeshR But 0 is not a value in the domain regardless, it is beyond question. – Jose M Serra Nov 22 '17 at 14:01
  • Wikipedia says that some people call removable singularities as discontinuities, and hence the confusion. Anyway, @RajeshR, the question is wrong anyway, because $f(t)$ is definitely not a function. A function comes with a domain and a set of mappings, and the question does not state a domain so it is wrong. Also, once you define the domain then it would be $f$ that is the function, not $f(t)$. – user21820 Dec 03 '17 at 10:03