Question

If the value of \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \left( a-n \right)nx-\tan x \right)\sin nx}{{{x}^{2}}}\] is equal to 0, where \[n\in R-\left\{ 0 \right\}\], then the value of ‘a’ is equal to
(a) 0
(b) \[\dfrac{n}{n+1}\]
(c) n
(d) \[n+\dfrac{1}{n}\]

Answer 1

Hint: In this question, first all multiply nx to both the numerator and denominator and use \[\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\sin a}{a}=1\]. Then separate the terms and use \[\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\tan a}{a}=1\] and then substitute the limit equal to 0 to find the value of a.

Complete step-by-step answer:
Here, we have to find the value of ‘a’ if \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \left( a-n \right)nx-\tan x \right)\sin nx}{{{x}^{2}}}\] is equal to 0, where \[n\in R-\left\{ 0 \right\}\]. Let us consider the limit given in the question.
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \left( a-n \right)nx-\tan x \right)\sin nx}{{{x}^{2}}}\]
By multiplying both numerator and denominator by nx in RHS of the above equation, we get,
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \left( a-n \right)nx-\tan x \right)\sin nx.\left( nx \right)}{{{x}^{2}}.\left( nx \right)}\]
We can also write the above equation as,
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \left( a-n \right)nx-\tan x \right).\left( nx \right)}{{{x}^{2}}}.\left( \dfrac{\sin nx}{nx} \right)\]
We know that \[\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\sin a}{a}=1\]. By using this in the above equation and canceling the like terms, we get,
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \left( a-n \right)nx-\tan x \right).n}{x}.1\]
We can also write the above equation as
\[L=\underset{x\to 0}{\mathop{\lim }}\,\left( \left( \dfrac{\left( a-n \right).{n}^{2}x}{x} \right)-\left( \dfrac{\tan x}{x} \right) \right).n\]
\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\text{ }{{n}^{2}}.\left( a-n \right)-\underset{x\to 0}{\mathop{\lim }}\,\text{ }n.\left( \dfrac{\tan x}{x} \right)\]
We know that \[\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\tan a}{a}=1\]. By using this in the above equation, we get,
\[L={{n}^{2}}\left( a-n \right)-n\]
We are given that L = 0, So by substituting it in the above equation, we get,
\[{{n}^{2}}\left( a-n \right)-n=0\]
\[\Rightarrow {{n}^{2}}\left( a-n \right)=n\]
\[\left( a-n \right)=\dfrac{n}{{{n}^{2}}}\]
\[\Rightarrow a-n=\dfrac{1}{n}\]
So, we get,
\[a=n+\dfrac{1}{n}\]
Hence, option (d) is the right answer.

Note: Students must remember the formula \[\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\sin a}{a}=1\] and \[\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\tan a}{a}=1\] as they are very useful while solving the question involving limits. Also, first of all, evaluate the whole limit and eliminate the x and then only substitute its value.