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

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.