lim n → ∞ 1 + 1 + 1 2 + … … + 1 n n 2 n is equal to

It is 1 ∞ form L = e l i m n → ∞ 1 + 1 2 + 1 3 + … . . + 1 n n L = e l i m n → ∞ 1 n + 1 2 n + 1 3 n + … . . + 1 n 2 ∴ L = e 0 + 0 + 0 + … . . + 0 ⇒ L = e 0 = 1

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Mathematics - Limits Question with Solution | TestHub

MathematicsLimits1 power infinty basedHard2 minPYQ_2021

MathematicsHardsingle choice

$\lim_{n \to \infty} {(1 + \frac{1 + \frac{1}{2} + \dots \dots + \frac{1}{n}}{n^{2}})}^{n}$ is equal to

Options:

Answer:

Solution:

It is $1^{\infty}$ form

$L = e^{\underset{n \to \infty}{l i m} (\frac{1 + \frac{1}{2} + \frac{1}{3} + \dots . . + \frac{1}{n})}{n})}$

$L = e^{\underset{n \to \infty}{l i m} (\frac{1}{n} + \frac{1}{2 n} + \frac{1}{3 n} + \dots . . + \frac{1}{n^{2}})}$

$∴ L = e^{(0 + 0 + 0 + \dots . . + 0)}$

$\Rightarrow L = e^{0} = 1$

Stream:JEESubject:MathematicsTopic:LimitsSubtopic:1 power infinty based

⏱ 2mℹ️ Source: PYQ_2021

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