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August 19th, 2019, 10:22 PM   #1
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Hard Limit

Evaluate the limit without L'hôpital’s rule.
$\displaystyle \lim_{y\rightarrow \infty }\frac{\ln(yn)}{y^n } \; $, $\displaystyle n\in \mathbb{N}.$

Last edited by skipjack; August 20th, 2019 at 03:05 AM.
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August 20th, 2019, 02:35 AM   #2
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$\displaystyle L=\lim_{y\rightarrow \infty} \frac{\ln(y)}{y^n } +\lim_{y\rightarrow \infty}\frac{\ln(n)}{y^n }=\lim_{y\rightarrow \infty}\frac{\ln(y)}{y^n }=\lim_{y\rightarrow \infty }\frac{\ln(e^y)}{e^{yn} }=\lim_{y\rightarrow \infty}n^{-1} \frac{y}{e^y}=\lim_{y\rightarrow \infty}(en)^{-1} \frac{1+y}{e^y }=(en)^{-1}L$

Now the equation is $\displaystyle L=(en)^{-1} L$. Since $\displaystyle y^{1-y} < \frac{y}{e^y } < \frac{y}{1+y} \; $ the limit converges .
$\displaystyle L=(ne)^{-1} L \: $ has solution L=0.

Last edited by idontknow; August 20th, 2019 at 03:15 AM.
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August 21st, 2019, 07:27 AM   #3
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Your $n^{-1}$ isn't correct when it appears.

The limit $\displaystyle \lim_{x \to \infty} \frac{\ln^a{x}}{x^b} = 0$ where $a > 0$ and $b > 0$ is a standard result that is often simply quoted. Whether you are allowed to do that or not depends on what you have been given (or proved) as results.

\begin{alignat}{2}
0 &< \frac1t &&< t^{c-1} &&& (c > 0,\, t> 1) \\
\int_1^x 0 \,\mathrm dt &< \int_1^x \frac1t \,\mathrm dt &&< \int_1^xt^{c-1}\,\mathrm dt &&&(x > 1) \\
0 &< \ln x &&< x^c - \frac1c &&< x^c \\
0 &< \frac{\ln^a x}{x^b} &&< x^{ac-b} &&& (a > 0, \, b > 0) \quad \text{raising to the power $a$ and dividing by $x^b$}\\
\end{alignat}
With $c=\frac{b}{2a} > 0$ we have $ac-b=-\frac{b}2 < 0$ and so, the limit of the left- and right-hand expressions as $x \to \infty$ is zero and thus the limit $$\lim_{x \to \infty} \frac{\ln^a{x}}{x^b} = 0 \qquad (a > 0, \, b > 0)$$
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Last edited by v8archie; August 21st, 2019 at 07:30 AM.
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