Limits, Continuity & Differentiability
General
Grade None
Question:
<p>If limn→∞
√
2n2 + n −λ
√
2n2 −n
=
1
√
2 (where λ is a real number), then:</p>
<p>λ = 1</p>
<p>λ = −1</p>
<p>λ = ±1</p>
<p>λ ∈(−∞, 1)</p>
Step-by-Step Solution
Key Concept: General
<p><strong>1</strong>: For the limit to be a finite non-zero value, the leading terms must cancel. Thus \lambda = 1</p> (since \lambda = -1 would lead to \infty).<p><strong>2</strong>: With \lambda = 1, rationalize:</p> lim n\to \infty (2n2 + n) -(2n2 -n) \sqrt 2n2 + n + \sqrt 2n2 -n = lim n\to \infty 2n n \sqrt 2 + n \sqrt 2 = 2 2 \sqrt 2 = 1 \sqrt 2<p><strong>3</strong>: This matches the given value, confirming \lambda = 1.</p>
Correct Answer: 1