Limit with sin^2x

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Limit with sin^2x

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Quantum_fluctuations

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Quantum_fluctuations

#1 h Apr 26, 2020, 2:30 AM

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Evaluate:

$\lim_{x \to 0} \left( 1^{1/\sin^2 x} + 2^{1/\sin^2 x} + 3^{1/\sin^2 x} + . . . + n^{1/\sin^2 x} \right)^{\sin^2 x}$

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#2 h Apr 26, 2020, 3:39 AM

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Hint: Take e^(ln (of that quantity))

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Quantum_fluctuations

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Quantum_fluctuations

#3 h Apr 26, 2020, 3:42 AM

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After that?

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#4 h Apr 26, 2020, 4:54 AM

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The natural logarithm of the quantity is then asymptotic to ${x^2\ln\left(\frac{n^{\left(\frac{1}{x^2}-1\right)}}{\frac{1}{x^2}-1}\right)}$ as $x\rightarrow 0$ .

From there it is easy to see that the given limit is $e^{\ln(n)}=n$ .

This post has been edited 2 times. Last edited by CALCMAN, Apr 26, 2020, 6:18 AM
Reason: We speak not of this incident.

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TuZo

#5 h Apr 26, 2020, 2:38 PM • 1 Y

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Solution:
Denote the limit $L$ , and we apply the logarithm: $\ln L=\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln \sum\limits_{k=1}^{n}{{{k}^{\frac{1}{{{\sin }^{2}}x}}}}}{\frac{1}{{{\sin }^{2}}x}}$ . Denote $\frac{1}{{{\sin }^{2}}x}=y\to \infty$ , we get $\ln L=\underset{y\to \infty }{\mathop{\lim }}\,\frac{\ln \sum\limits_{k=1}^{n}{{{k}^{y}}}}{y}\overbrace{=}^{l'Hospital}\underset{y\to \infty }{\mathop{\lim }}\,\frac{{{1}^{y}}\ln 1+{{2}^{y}}\ln 2+...+{{n}^{y}}\ln n}{{{1}^{y}}+{{2}^{y}}+...+{{n}^{y}}}=\underset{y\to \infty }{\mathop{\lim }}\,\frac{{{n}^{y}}\left( {{\left( \frac{1}{n} \right)}^{y}}\ln 1+{{\left( \frac{2}{n} \right)}^{y}}\ln 2+...+{{\left( \frac{n-1}{n} \right)}^{y}}\ln (n-1)+\ln n \right)}{{{n}^{y}}\left( {{\left( \frac{1}{n} \right)}^{y}}+{{\left( \frac{2}{n} \right)}^{y}}+...+{{\left( \frac{n-1}{n} \right)}^{y}}+\operatorname{l} \right)}=\ln n$ . So $\ln L=\ln n\Rightarrow L=n$ .

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Figaro

#6 h Apr 27, 2020, 8:07 AM • 1 Y

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Let $t=\frac{1}{\sin^2{x}}$ . The limit becomes $\lim_{t \to \infty} \left( 1^t+2^t+3^t+...+n^t\right)^{1/t}$ . We see that

$\lim_{t \to \infty} \left( n^t\right)^{1/t}\leq\lim_{t \to \infty} \left( 1^t+2^t+3^t+...+n^t\right)^{1/t}\leq\lim_{t \to \infty} \left( n^t+n^t+n^t+...+n^t\right)^{1/t}$

and the squeeze theorem shows that the limit is $n$ .

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Quantum_fluctuations

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Quantum_fluctuations

#7 h Apr 27, 2020, 3:48 PM

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Figaro wrote:

Let $t=\frac{1}{\sin^2{x}}$ . The limit becomes $\lim_{t \to \infty} \left( 1^t+2^t+3^t+...+n^t\right)^{1/t}$ . We see that

$\lim_{t \to \infty} \left( n^t\right)^{1/t}\leq\lim_{t \to \infty} \left( 1^t+2^t+3^t+...+n^t\right)^{1/t}\leq\lim_{t \to \infty} \left( n^t+n^t+n^t+...+n^t\right)^{1/t}$

and the squeeze theorem shows that the limit is $n$ .

Interesting ... :-D

:-D

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#8 h Apr 23, 2025, 7:25 AM

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$L = \lim_{x \to 0} \left( 1^{1/\sin^2 x} + 2^{1/\sin^2 x} + 3^{1/\sin^2 x} + . . . + n^{1/\sin^2 x} \right)^{\sin^2 x}$

Claim $:- \lim_{n \to \infty} (a^n + b^n)^{1/n} = max(a,b)$

Proof $:- WLOG$ let $a > b$ then we've

$\lim_{n \to \infty} (a^n + b^n)^{1/n} = \lim_{n \to \infty} a\left(1 + \left(\frac{b}{a}\right)^n\right)^{1/n} = a = max(a,b)$

Therefore, $L = max(1,2,....,n) = \boxed{n}$

This post has been edited 9 times. Last edited by P162008, Apr 23, 2025, 7:41 AM
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