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At $t=0$, the function $f(t)=\frac{\sin t}{t}$ has

(A) a minimum

(B) a discontinuity

(C) a point of inflection

(D) a maximum
asked in Calculus by Loyal (7.4k points)
edited by | 992 views

2 Answers

+2 votes
Best answer

$$\begin{align*} \lim_{t \rightarrow 0} \frac{\sin t}{t} &= \lim_{t \rightarrow 0} \frac{\cos t}{1} \qquad \text{;using L'Hospital's Rule for } \frac{0}{0} \text{ form}\\ &= 1 \end{align*}$$

visually we can see clearly for the plot $y = \frac{\sin x}{x}$

that data is about the neighbourhood of function at $x=0$

function is not defined at x=0 so, this means we are talking about the point which is not in the function domain hence, we cannot say anything about it.

answered by Boss (30.5k points)
edited by
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Amar, but I think LH's rule is applicable only when t → 0

So, I agree that L.H.L = R.H.L. = 1.

but at t = 0 , how can we say that (sint)/t  =1 ?
0 votes
For any given function,if anything like continuity,discontinuity,maxima,minima are asked, it means that we have to consider in its domain.

here domain  =  R-{0}.

Hence function is not defined at x = 0 at all. So we can't answer any question related to x = 0;

But yes when x->0, function is defined and its limiting value exists and it equals to '1'.

At x -> 0, functional gives maximum value also.

But at x=0; nothing can be said.

So if we defined function at x= 0 as f(0) = 1,then we can say function is continuous & maximum value exists at x=0.

So don't worry,IIT will never ask such confusing questions..just move on.
answered by Active (3.5k points)
0
So don't worry,IIT will never ask such confusing questions..just move on.

Was asked in GATE 2010 for 2 marks.. XD

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