Question

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

Answer 1

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:
 $ \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
 $ \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
 $ \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{x\left( {x + 1} \right)}}{{x + 1}} + \dfrac{1}{{x + 1}} - ax - b} \right) = 4 $
 $ \left( {x + 1} \right) $ Gets cancelled
 $
  \mathop {\lim }\limits_{x \to \infty } \left( {x + \dfrac{1}{{x + 1}} - ax - b} \right) = 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
 $ \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.
 $
  \left( {1 - a} \right) = 0 \\
  a = 1 \\
  $
Now, equation (4) becomes,
 $ \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{x + 1}} - b} \right) = 4 $
Taking x common form denominator,
 $ \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, $ \dfrac{1}{x} $ must be equal to zero.
Therefore, equation (5) becomes,
 $
   - b = 4 \\
  b = - 4 \;
  $
Hence, our answer is option B) $ a = 1,b = - 4 $
So, the correct answer is “Option B”.

Note: We can also solve this question using L-Hospital method.
 $ \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right) = 4 $
Taking L.C.M
 $ \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.
\[
  \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{2x + 1 - 2ax - a - b}}{1}} \right) = 4 \\
  \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{\left( {2 - 2a} \right)x + \left( {1 - a - b} \right)}}{1}} \right) = 4 \;
 \]
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.
 $ \Rightarrow 2 - 2a = 0 $ - - - - - (1)
 $ \Rightarrow 1 - a - b = 4 $ - - - - - - (2)
By solving equations (1) and (2), we get
 $ a = 1,b = - 4 $