The value of n-lim1n-r - 1r - 2nr-n2 - r2 equals-

The correct option is B √(5) 1 n→∞lim1n∑_r = 1^r = 2nrn√(1 + (rn )^2) = ∫_0^2x√(1 + x^2)dx = (√(1 + x^2))_0^2 = √(5) 1 Hence choice (b) is c ...

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Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

The value of ...

Question

The value of $lim n \to \infty \frac{1}{n} r = 2 n \sum r = 1 \frac{r}{\sqrt{n^{2} + r^{2}}}$ equals.

\sqrt{5} + 1

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\sqrt{5} - 1

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1 = \sqrt{5}

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None of these

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Solution

The correct option is B $\sqrt{5} - 1$
$lim n \to \infty \frac{1}{n} r = 2 n \sum r = 1 \frac{\frac{r}{n}}{\sqrt{1 + {(\frac{r}{n})}^{2}}}$
$= \int_{0}^{2} \frac{x}{\sqrt{1 + x^{2}}} d x = (\sqrt{1 + x^{2}})_{0}^{2} = \sqrt{5} - 1$
Hence choice (b) is correct answer.

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The value of limn→∞1nr=2n∑r=1r√n2+r2 equals.

The correct option is B √5−1limn→∞1nr=2n∑r=1rn√1+(rn)2=∫20x√1+x2dx=(√1+x2)20=√5−1Hence choice (b) is correct answer.

The value of $lim n \to \infty \frac{1}{n} r = 2 n \sum r = 1 \frac{r}{\sqrt{n^{2} + r^{2}}}$ equals.

The correct option is B $\sqrt{5} - 1$
$lim n \to \infty \frac{1}{n} r = 2 n \sum r = 1 \frac{\frac{r}{n}}{\sqrt{1 + {(\frac{r}{n})}^{2}}}$
$= \int_{0}^{2} \frac{x}{\sqrt{1 + x^{2}}} d x = (\sqrt{1 + x^{2}})_{0}^{2} = \sqrt{5} - 1$
Hence choice (b) is correct answer.