Hard Limit

Dec 2015
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Evaluate the limit without L'hôpital’s rule.
\(\displaystyle \lim_{y\rightarrow \infty }\frac{\ln(yn)}{y^n } \; \), \(\displaystyle n\in \mathbb{N}.\)
 
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Dec 2015
1,076
166
Earth
\(\displaystyle L=\lim_{y\rightarrow \infty} \frac{\ln(y)}{y^n } +\lim_{y\rightarrow \infty}\frac{\ln(n)}{y^n }=\lim_{y\rightarrow \infty}\frac{\ln(y)}{y^n }=\lim_{y\rightarrow \infty }\frac{\ln(e^y)}{e^{yn} }=\lim_{y\rightarrow \infty}n^{-1} \frac{y}{e^y}=\lim_{y\rightarrow \infty}(en)^{-1} \frac{1+y}{e^y }=(en)^{-1}L\)

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.
 
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v8archie

Math Team
Dec 2013
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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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