Question

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Answer 1

Answer:

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Answer 2

Solution :-

$\displaystyle{ \lim_{n \to \infty} \left( \dfrac{x^2 + x + 1}{x+ 1} - ax-b \right) }= 4$

By further solving :-

$\displaystyle{ \lim_{n \to \infty} \left( \dfrac{x^2 + x + 1- ax^2 - bx -ax -b }{x+ 1} \right) } = 4\\\\\displaystyle{ \lim_{n \to \infty} \left( \dfrac{x^2 (1 -a)+ x (1-a-b)+ 1-b}{x+ 1} \right) }=4$

By dividing numerator and denominator by x

$\displaystyle{ \lim_{n \to \infty} \left( \dfrac{\dfrac{x^2}{x}(1 -a)+ \dfrac{x}{x}(1-a-b)+\dfrac{ 1-b}{x}}{\dfrac{x+ 1}{x}}\right) }=4 \\\\ \displaystyle{ \lim_{n \to \infty} \left( \dfrac{x(1 -a)+ (1-a-b)+ \dfrac{1-b}{x}}{1 + \dfrac{1}{x}} \right) }= 4$

Now to limit to be exist :-

→ 1 - a = 0

→ a = 1

Now further solving :-

$\displaystyle{ \lim_{n \to \infty} \left( \dfrac{x(1 - 1)+ (1-1-b)+ \dfrac{1-b}{x}}{1 + \dfrac{1}{x}} \right) }= 4\\\\\displaystyle{ \lim_{n \to \infty} \left( \dfrac{ (-b)+ \dfrac{1-b}{x}}{1 + \dfrac{1}{x}} \right) }= 4$

Now by applying limit

$\displaystyle{ \lim_{n \to \infty} \left( \dfrac{ (-b)+ \dfrac{1-b}{x}}{1 + \dfrac{1}{x}} \right)=4}$

$\implies \dfrac{ -b+ \dfrac{1-b}{\infty}}{1 + \dfrac{1}{\infty}}= 4\\\\\implies \dfrac{-b}{1} = 4 \\\\\implies b = -4$

So

a = 1

b = -4

Option (b) is the answer