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+3 votes
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 Boss (7k points)  
edited by | 248 views

2 Answers

+3 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 Veteran (27.5k points)  
edited by
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 Loyal (3.4k points)  
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