Question

solve this
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Answer 1

$\Huge\boxed{\tt\green{✠Answer}}$

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$\pmb{\sf{\gray{According \: to \: the \: Question }}}$

$\bold{ \gamma = \displaystyle \lim_{x \to \infty} \Bigg(\sum\limits_{k=1}^{n} \frac{1}{k} - \displaystyle \ln \: n \Bigg)}$

$\longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg(1 + \frac{1}{2} + \frac{1}{3}. \: . \: . \: . \frac{1}{n} - \displaystyle \ln \: n \bigg)}$

$\pmb{\sf{\gray{ Multiply \: and \: divide \: 'n' \:,}}} \\ \pmb{\sf{\gray{denominator \: and \: numerator }}}$

$\longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg( \frac{n}{1n} + \frac{n}{2n} + \frac{n}{3n} ... \: ... + \frac{n}{n^{2} } - \displaystyle \frac{n \: \ln \: n}{n} \bigg)}$

$\pmb{\sf{\gray{By \: taking \: in \: common }}}$

$\longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg( \frac{1}{n} \bigg( \frac{n}{1} + \frac{n}{2} + \frac{n}{3} ... \: ... + \frac{n}{n } - \displaystyle n \: \ln \: n \bigg)}$

$\tt{So,}\longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg( \frac{1}{x} \bigg( \frac{x}{1} + \frac{x}{2} + \frac{x}{3} ... \: ... + 1 - \displaystyle x\: \ln \:x\bigg)}$

$\pmb{\sf{\gray{Now \: given \: limit \: x \longrightarrow \infty \: by \: putting,}}}$

$\tt{So,} \longrightarrow\bold{ \gamma = \displaystyle \lim_{x \to \infty} \bigg( \frac{1}{ \infty } \bigg( \frac{ \infty }{1} + \frac{ \infty }{2} + \frac{ \infty }{3} ... \: ... + 1 - \displaystyle \infty \: \ln \: \infty \bigg)}</p><p>$

$\sf{ \longrightarrow \infty = \frac{1}{ \infty } \infty = \frac{ \infty }{ \infty } }$

$\sf{ \longrightarrow \infty = \frac{1}{ \infty } \infty = \frac{ \bcancel\infty }{ \bcancel{\infty } }}$

$\huge \boxed{ \tt{ \gamma = 0}}$

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Answer 2

Answer:

From the NCERT Solutions at BYJU’S, students will clearly learn about all the chemical reactions which occur in our day-to-day activities. The solutions explain each and every minute concept in the best way possible so that students do not face any problem in the exam. It not only boosts their exam preparation, but also provides a strong foundation of fundamental concepts, which frequently appear in various competitive exams. Using these solutions, students will understand how to approach complex questions that would appear in the exam and answer them with confidence.

Download NCERT Solutions for Class 11 Chemistry PDF to revise all the concepts.

Step-by-step explanation:

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