Find the value of the expression given below,
\[\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{-3n+{{(-1)}^{n}}}{4n-{{(-1)}^{n}}},(n\in N)\]
(a)$-\dfrac{3}{4}$
(b)0 if n is even.
(c)$-\dfrac{3}{4}~$ If ‘n’ is odd
(d)None of this

Hint: To solve the above problem just take ‘n’ common from both numerator and denominator. Then apply the limits to get the desired result.

Complete step-by-step answer:
Firstly we will write the expression given in the problem and assume it as ‘L’
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{-3n+{{(-1)}^{n}}}{4n-{{(-1)}^{n}}}$
If we observe the problem carefully, we will come to know that the term ${{(-1)}^{n}}$ takes two values, one when n is even and another value when n is odd. Therefore, we have to solve this problem in two cases which are as follows,
Case 1, when ‘n’ is even.
As we all know the value of ${{(-1)}^{n}}$if ‘n’ even comes +1.
By substituting the above value in ‘L’ we will get,
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{-3n+1}{4n-1}$
Now to solve further just take ‘n’ common from both numerator and denominator, therefore we will get,
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{n\left( -3+\dfrac{1}{n} \right)}{n\left( 4-\dfrac{1}{n} \right)}$
We can easily see that ‘n’ can be cancelled out from numerator and denominator, therefore we will get,
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( -3+\dfrac{1}{n} \right)}{\left( 4-\dfrac{1}{n} \right)}$
Now we will just put the limits to get the final answer,
$\therefore L=\dfrac{\left( -3+\dfrac{1}{\infty } \right)}{\left( 4-\dfrac{1}{\infty } \right)}$
As we all know that the value of $\dfrac{1}{\infty }$is tending to Zero, therefore we will get,
$\therefore L=\dfrac{\left( -3+0 \right)}{\left( 4-0 \right)}$
$\therefore L=-\dfrac{3}{4}$………………………………… (1)
Case 2, when ‘n’ is odd.
As we all know the value of ${{(-1)}^{n}}$if ‘n’ even comes -1.
By substituting the above value in ‘L’ we will get,
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{-3n-1}{4n-\left( -1 \right)}$
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{-3n-1}{4n+1}$
Now to solve further just take ‘n’ common from both numerator and denominator, therefore we will get,
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{n\left( -3-\dfrac{1}{n} \right)}{n\left( 4+\dfrac{1}{n} \right)}$
We can easily see that ‘n’ can be cancelled out from numerator and denominator, therefore we will get,
$\therefore L=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( -3-\dfrac{1}{n} \right)}{\left( 4+\dfrac{1}{n} \right)}$
Now we will just put the limits to get the final answer,
$\therefore L=\dfrac{\left( -3-\dfrac{1}{\infty } \right)}{\left( 4+\dfrac{1}{\infty } \right)}$
As we all know that the value of $\dfrac{1}{\infty }$is tending to Zero, therefore we will get,
$\therefore L=\dfrac{\left( -3-0 \right)}{\left( 4+0 \right)}$
$\therefore L=-\dfrac{3}{4}$………………………………… (2)
From (1) and (2) we can say that,
$L=-\dfrac{3}{4}$ $n\in N$
Therefore we have the answer i. e. the value of the given expression is $-\dfrac{3}{4}$ for$(n\in N)$.
Hence, the correct answer is option (a).

Note: We can solve this problem by using the L-Hospital’s Rule directly which will save our time too, but we have to solve this using both cases as there are chances of silly mistakes.
L-Hospital’s Rule:
\[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]