Question

Needhdjdjdjdjdjdnhiskdjdjdjdjdjdjjshdjdjdjdjdjdjdjd​

Answer 1

9514 1404 393

Answer:

  -(√2)/2

Step-by-step explanation:

The expression evaluated at n=a gives the indeterminate form 0/0, so L'Hopital's rule can be used to find the limit. The second expression comes from differentiating numerator and denominator. Then the form with n=a is no longer indeterminate.

  $\displaystyle\lim_{n\to a}{\frac{\sqrt{2n}-\sqrt{3n-a}}{\sqrt{n}-\sqrt{a}}}=\lim_{n\to a}{\frac{\frac{2}{2\sqrt{2n}}-\frac{3}{2\sqrt{3n-a}}}{\frac{1}{2\sqrt{n}}-0}}\\\\=\sqrt{a}\left(\frac{2}{\sqrt{2a}}-\frac{3}{\sqrt{3a-a}}}\right)=\boxed{-\frac{1}{\sqrt{2}}}$