If \mathop \lim \limits_n \to \infty ( \sqrt n^2 - n - 1 + n\alpha + \beta ) = 0 , then 8(\alpha+\beta) is equal to :

Limits Continuity and Differentiability PYQ — JEE Main 2022 · 25 Jul · Shift 1 · Q25 with Solution

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Correct answer: C

We need $\lim_{n\to\infty}\left(\sqrt{n^2-n-1}+n\alpha+\beta\right)=0.$

We must choose $\alpha,\beta$ so that the expression tends to $0$ .

Expand $\sqrt{n^2-n-1}$ for large $n$ .

Write $\sqrt{n^2-n-1}=n\sqrt{1-\frac1n-\frac1{n^2}}.$

Using $\sqrt{1+x}=1+\frac{x}{2}+o(x) \quad (x\to 0),$ with $x=-\frac1n-\frac1{n^2},$ we get $\sqrt{1-\frac1n-\frac1{n^2}}=1-\frac{1}{2n}+o\left(\frac1n\right).$

Hence $\sqrt{n^2-n-1}=n\left(1-\frac{1}{2n}+o\left(\frac1n\right)\right)=n-\frac12+o(1).$

Substitute into the given expression:

=\left(n-\frac12+o(1)\right)+n\alpha+\beta.$$ So $$=n(1+\alpha)+\left(\beta-\frac12\right)+o(1).$$ For the limit to be $0$, both the coefficient of $n$ and the constant term must vanish. Thus, $$1+\alpha=0 \implies \alpha=-1,$$ and $$\beta-\frac12=0 \implies \beta=\frac12.$$ 4. Now compute: $$\alpha+\beta=-1+\frac12=-\frac12.$$ Therefore, $$8(\alpha+\beta)=8\left(-\frac12\right)=-4.$$ 5. Checking options: - A: $4$ ❌ - B: $-8$ ❌ - C: $-4$ ✅ - D: $8$ ❌ So the correct answer is **Option C**.

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View written solutionFree

Correct answer: C

We need $\lim_{n\to\infty}\left(\sqrt{n^2-n-1}+n\alpha+\beta\right)=0.$

We must choose $\alpha,\beta$ so that the expression tends to $0$ .

Expand $\sqrt{n^2-n-1}$ for large $n$ .

Write $\sqrt{n^2-n-1}=n\sqrt{1-\frac1n-\frac1{n^2}}.$

Using $\sqrt{1+x}=1+\frac{x}{2}+o(x) \quad (x\to 0),$ with $x=-\frac1n-\frac1{n^2},$ we get $\sqrt{1-\frac1n-\frac1{n^2}}=1-\frac{1}{2n}+o\left(\frac1n\right).$

Hence $\sqrt{n^2-n-1}=n\left(1-\frac{1}{2n}+o\left(\frac1n\right)\right)=n-\frac12+o(1).$

Substitute into the given expression:

=\left(n-\frac12+o(1)\right)+n\alpha+\beta.$$ So $$=n(1+\alpha)+\left(\beta-\frac12\right)+o(1).$$ For the limit to be $0$, both the coefficient of $n$ and the constant term must vanish. Thus, $$1+\alpha=0 \implies \alpha=-1,$$ and $$\beta-\frac12=0 \implies \beta=\frac12.$$ 4. Now compute: $$\alpha+\beta=-1+\frac12=-\frac12.$$ Therefore, $$8(\alpha+\beta)=8\left(-\frac12\right)=-4.$$ 5. Checking options: - A: $4$ ❌ - B: $-8$ ❌ - C: $-4$ ✅ - D: $8$ ❌ So the correct answer is **Option C**.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

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