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Calculate the limit:

$ \lim_{x\to\infty}\frac{\ln(x^\frac{5}2+7)}{x^2} $

This is a part of the whole limit that I'm trying to calculate, but it is this part I have a hard time to figure out why this limit is zero.

Any ideas? Is it reasonable to say that the quotient will be zero because of the denominator function grows faster than the nominator?

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    By what I wrote, if $x\gt 10$ (we can use a cheaper $x$, but that doesn't matter) your function is between $0$ and $\frac{\ln 2+(3/2)\ln x}{x^2}$. Fairly easily this goes to $0$, so by Squeezing our function does.2012-10-02

3 Answers 3

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You probably know that $e^x\ge x+1$ and that is all you need.

Letting $x=\ln y$ with $y>0$ this becomes $y\ge \ln(y) +1$ or $\ln(y)\le1-y\mathrm{\quad for\ }y>0.$

For $x>7^{\frac25}$ we therefore have $\ln(x^{\frac52}+7)<\ln(2x^{\frac52})=\ln2+\frac52\ln x<1+\frac52(x-1)<\frac52 x.$ This makes $0<\frac{\ln(x^{\frac52}+7)}{x^2}<\frac5{2x}\to 0$ if $x>7^{\frac25}$.

Do you see how this generalizes to $\frac{\ln(p(x))}{q(x)}\to0$ for arbitrary polynomials $p(x)>0,q(x)\ne0$?

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Hint: in this case,

$\lim_{x\to\infty} \frac{f(x)}{h(x)}=\lim_{x\to\infty} \frac{f'(x)}{h'(x)}$

Here for more info

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    Thanks Julien. Although we haven't been through L'Hospital rule yet in our lectures, it seems useful.2012-10-02
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You don't really need L'Hopitals rule to answer this question. All you need to know is that the ln(x) function grows much slower than x, and the answer will be obvious from that.

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    Polynomial growth dominates logarithmic growth in even the most extreme cases.2012-10-02