1.If limn→∞(2n2+n−2n2−n)=1λ\displaystyle \lim_{n \to \infty} \left(\sqrt{2n^2 + n} - \sqrt{2n^2 - n}\right) = \frac{1}{\sqrt{\lambda}}n→∞lim(2n2+n−2n2−n)=λ1 (where λ\lambdaλ is a real number), then λ\lambdaλ is equal toa.λ=1\lambda = 1λ=1b.λ=−1\lambda = -1λ=−1c.λ=±1\lambda = \pm 1λ=±1d.λ∈(−∞,1)\lambda \in (-\infty, 1)λ∈(−∞,1)Login to continueOnly logged in users canattempt or see the solution.
