If $$\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\left( {a - n} \right)nx - \tan \,x} \right)\sin \,nx}}{{{x^2}}} = 0$$ where $$n$$ is nonzero real number, then $$a$$ is equal to-
A.
$$0$$
B.
$$\frac{{n + 1}}{n}$$
C.
$$n$$
D.
$$n + \frac{1}{n}$$
Answer :
$$n + \frac{1}{n}$$
Solution :
We are given that
$$\mathop {\lim }\limits_{x \to 0} \frac{{\left[ {\left( {a - n} \right)nx - \tan \,x} \right]\sin \,nx}}{{{x^2}}} = 0$$
where $$n$$ is non zero real number
$$\eqalign{
& \Rightarrow \mathop {\lim }\limits_{x \to 0} n.\frac{{\sin \,nx}}{{nx}}\left[ {\left\{ {\left( {a - n} \right)n - \frac{{\tan \,x}}{x}} \right\}} \right] = 0 \cr
& \Rightarrow 1.n\left[ {\left( {a - n} \right)n - 1} \right] = 0 \cr
& \Rightarrow a = \frac{1}{n} + n\,\,{\text{or}}\,\,\,n + \frac{1}{n} \cr} $$
Releted MCQ Question on Calculus >> Limits
Releted Question 1
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If $$\eqalign{
& f\left( x \right) = \frac{{\sin \left[ x \right]}}{{\left[ x \right]}},\,\,\left[ x \right] \ne 0 \cr
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