MCQ
$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $ equals
- A$1 + \sqrt 5 $
- ✓$ - 1 + \sqrt 5 $
- C$ - 1 + \sqrt 2 $
- D$1 + \sqrt 2 $
✓
Answer
Correct option: B.
$ - 1 + \sqrt 5 $
b
(b) $L = \mathop {\lim }\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^{2n} {} \frac{1}{n}\,.\,\frac{{r/n}}{{\sqrt {1 + {{(r/n)}^2}} }}$
(b) $L = \mathop {\lim }\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^{2n} {} \frac{1}{n}\,.\,\frac{{r/n}}{{\sqrt {1 + {{(r/n)}^2}} }}$
$= \int_0^2 {} \frac{x}{{\sqrt {1 + {x^2}} }}\,dx = \sqrt 5 - 1$.
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