N-lim -1-n2-12-1-n2-22-1-2n-1-

The correct option is C π/2 lt_n→∞1/n [ 1/√(1+(^1/_n)^2)+1/√(1+(^2/_n)^2)+......... ] lt_n →∞ 1/n ∑_r=1^n 1/√(1+(^2/_n)^2) ⇒∫_0^11/√(1 x^ ...

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Byju's Answer

Standard XII

Mathematics

Definite Integral as Limit of Sum

n→∞lim [1/√n2...

Question

$lim n \to \infty [\frac{1}{\sqrt{n^{2} - 1^{2}}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + \dots + \frac{1}{\sqrt{2 n - 1}}] =$

π

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2 π

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π / 2

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3 π / 2

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Solution

The correct option is C $π / 2$
$l t_{n \to \infty} \frac{1}{n} [\frac{1}{\sqrt{1 + (^{1} /_{n})^{2}}} + \frac{1}{\sqrt{1 + (^{2} /_{n})^{2}}} + . . . . . . . . .]$
$l t_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{\sqrt{1 + (^{2} /_{n})^{2}}}$
$\Rightarrow \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} d x = s i n^{- 1} (x) |_{0}^{1} =^{π} /_{2}$

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is equal to :

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limn→∞[1√n2−12+1√n2−22+…+1√2n−1]=

The correct option is C π/2ltn→∞1n[1√1+(1/n)2+1√1+(2/n)2+.........]ltn→∞ 1n ∑nr=1 1√1+(2/n)2⇒∫101√1−x2dx=sin−1(x)|10=π/2

$lim n \to \infty [\frac{1}{\sqrt{n^{2} - 1^{2}}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + \dots + \frac{1}{\sqrt{2 n - 1}}] =$