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Question: If \( \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \rig...

If lim⁡x→∞(x2+x+1x+1−ax−b)=4\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right) = 4 , then
A. a=1,b=4a = 1,b = 4
B. a=1,b=−4a = 1,b = - 4
C. a=2,b=−3a = 2,b = - 3
D. a=2,b=3a = 2,b = 3

Explanation

Solution

Hint : In this question, first of all simplify the equation a little and then as the limit is infinite, take the coefficient of x equal to zero and the remaining term equal to 4. This will give us the values of a and b.

Complete step by step solution:
Given expression:
lim⁡x→∞(x2+x+1x+1−ax−b)=4\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right) = 4 - - - - - - - - (1)
And we are supposed to find the values of a and b.
Taking x common in numerator, equation (1) becomes
lim⁡x→∞(x(x+1)+1x+1−ax−b)=4\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{x\left( {x + 1} \right) + 1}}{{x + 1}} - ax - b} \right) = 4 - - - - - - - - - - (2)
Separating numerator, equation (2) becomes
lim⁡x→∞(x(x+1)x+1+1x+1−ax−b)=4\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{x\left( {x + 1} \right)}}{{x + 1}} + \dfrac{1}{{x + 1}} - ax - b} \right) = 4
(x+1)\left( {x + 1} \right) Gets cancelled
lim⁡x→∞(x+1x+1−ax−b)=4   \mathop {\lim }\limits_{x \to \infty } \left( {x + \dfrac{1}{{x + 1}} - ax - b} \right) = 4 \\\ \\\
lim⁡x→∞(x−ax+1x+1−b)=4\mathop {\lim }\limits_{x \to \infty } \left( {x - ax + \dfrac{1}{{x + 1}} - b} \right) = 4 - - - - - - - - (3)
Taking x common in above equation, we get
lim⁡x→∞(x(1−a)+1x+1−b)=4\mathop {\lim }\limits_{x \to \infty } \left( {x\left( {1 - a} \right) + \dfrac{1}{{x + 1}} - b} \right) = 4 - - - - - - - - - (4)
Since the limit is infinite, the coefficient of x in equation (4) must be equal to zero.
(1−a)=0 a=1  \left( {1 - a} \right) = 0 \\\ a = 1 \\\
Now, equation (4) becomes,
lim⁡x→∞(1x+1−b)=4\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{x + 1}} - b} \right) = 4
Taking x common form denominator,
lim⁡x→∞(1x(1+1x)−b)=4\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{x\left( {1 + \dfrac{1}{x}} \right)}} - b} \right) = 4 - - - - - - - - - - (5)
Now, since the limit is infinite, 1x\dfrac{1}{x} must be equal to zero.
Therefore, equation (5) becomes,
\-b=4 b=−4   \- b = 4 \\\ b = - 4 \;
Hence, our answer is option B) a=1,b=−4a = 1,b = - 4
So, the correct answer is “Option B”.

Note : We can also solve this question using L-Hospital method.
lim⁡x→∞(x2+x+1x+1−ax−b)=4\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right) = 4
Taking L.C.M
lim⁡x→∞(x2+x+1−ax2−ax−bx−bx+1)=4\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + x + 1 - a{x^2} - ax - bx - b}}{{x + 1}}} \right) = 4
If we put the value of the limit in the above equation, the answer would be ∞∞\dfrac{\infty }{\infty } .
So, we take the derivative of the above expression.

lim⁡x→∞(2x+1−2ax−a−b1)=4 lim⁡x→∞((2−2a)x+(1−a−b)1)=4   \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{2x + 1 - 2ax - a - b}}{1}} \right) = 4 \\\ \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{\left( {2 - 2a} \right)x + \left( {1 - a - b} \right)}}{1}} \right) = 4 \;

Now, if we put the value of limit in the above equation, the answer would be ∞\infty .
But, the answer is 4.
So, the coefficient of x must be equal to 0 and the remaining term must be equal to 4.
⇒2−2a=0\Rightarrow 2 - 2a = 0 - - - - - (1)
⇒1−a−b=4\Rightarrow 1 - a - b = 4 - - - - - - (2)
By solving equations (1) and (2), we get
a=1,b=−4a = 1,b = - 4