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Question: If [] denotes the greatest integer function, for n\(\in \)N, then \(\underset{x\to 0}{\mathop{\lim }...

If [] denotes the greatest integer function, for n∈\in N, then lim⁡x→0 [nsin⁡xx]\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{n\sin x}{x} \right] is?
(a) n
(b) 0
(c) n-1
(d) n+1

Explanation

Solution

- Hint: To solve this problem, we should know the Taylor series expansion of sinx. This is given by-
sinx = x−x33!+x55!+....x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+....
We plug this in place of sin x and then continue to solve the problem.

Complete step-by-step solution -

We have,
lim⁡x→0 [nsin⁡xx]\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{n\sin x}{x} \right]
Now, to putting the Taylor series expansion of sinx in this expression, we get,
=lim⁡x→0 [n(x−x33!+x55!+....x)]\underset{x\to 0}{\mathop{\lim }}\,\left[ n\left( \dfrac{x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+....}{x} \right) \right]
After dividing all the terms by x, we get-
= lim⁡x→0 [n(1−x23!+x45!+....)]\underset{x\to 0}{\mathop{\lim }}\,\left[ n\left( 1-\dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+.... \right) \right]
Now, we re-arrange the terms slightly to get-
= lim⁡x→0 [n−n(x23!+x45!+....)]\underset{x\to 0}{\mathop{\lim }}\,\left[ n-n\left( \dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+.... \right) \right] -- (1)
Now, we analyse the term n(x23!+x45!+....)n\left( \dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+.... \right) in a bit more detail.
We see that as x→\to 0, n\left( \dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+.... \right)$$$\to $0. However, it will not be completely zero since x only tends to zero and is not actually zero. Further, since this expression has all square terms, it will not matter whether we approach form left hand limit of zero or the right hand limit of zero. This is because in both cases finally the term would eventually be positive due to the square terms. Further, since this representation of sinx is in the form of polynomial, we can safely say that it is continuous and differentiable everywhere. Now, we come back to analysing the term n\left( \dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+.... \right). Thus, for very small values of x(since x$\to $0), n\left( \dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+.... \right)willbeslightlygreaterthan0(atthesametimeverycloseto0).Now,wecomebacktosolving(1),weget,will be slightly greater than 0 (at the same time very close to 0). Now, we come back to solving (1), we get, \underset{x\to 0}{\mathop{\lim }},\left[ n-n\left( \dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+.... \right) \right]= =\underset{x\to 0}{\mathop{\lim }},\left[ n-a \right]Where,aisthevalueof Where, a is the value ofn\left( \dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+.... \right)$$ and will be slightly greater than 0.
=n-1
(Say a=0.001, [n-0.001] = n. This, we get from the definition of greatest integer function)
Hence, the correct answer is (c) n-1.

Note: Generally, to solve questions related to the greatest integer function involving trigonometric functions, one must be aware about the Taylor series expansion and sandwich theorem. Although, in this case sandwich theorem is not useful, it greatly simplifies our work in most cases. In all other cases, Taylor series expansion helps in solving the question (although, sometimes it is more tedious).