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MHTCET 18 Apr 2026 Shift 1 Paper

Question : 109 of 150

Marks: +1, -0

\lim _{x \rightarrow \infty}\left[\frac{x^{2}+x+1}{x+1}-a x-b\right]=3

a-b=\cdots

Solution:

Concept:

For the limit to exist as a finite number, the coefficient of

x

in the expression must be zero.

Explanation:

Divide

x^2+x+1

x+1

\frac{x^2+x+1}{x+1} = x + \frac{1}{x+1}

Substitute this into the given limit:

\lim_{x \to \infty}\left[(1-a)x + \frac{1}{x+1} - b\right] = 3

For a finite limit, the coefficient of

x

must vanish:

1-a = 0 \quad \Rightarrow \quad a = 1

x \to \infty

, we have

\frac{1}{x+1} \to 0

So the limit becomes

-b = 3

, giving

b = -3

Therefore,

a - b = 1 - (-3) = 4

Answer:

Option D:

4

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