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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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1  
They may want more. I would say that if $x\gt 10$, then $\ln(x^{3/2}+7)\lt \ln(2x^{3/2})=\ln 2+(3/2)\ln x$. Now you can use standard facts. –  André Nicolas Oct 2 '12 at 17:20
    
Hm, I don't follow... –  Curtain Oct 2 '12 at 17:29
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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. –  André Nicolas Oct 2 '12 at 17:33

3 Answers 3

up vote 1 down vote accepted

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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That would be a nice way to explain this limit. But why does it equal that in this case? –  Curtain Oct 2 '12 at 17:20
    
@JulianAssange See edit. –  user39572 Oct 2 '12 at 17:21
    
Thanks Julien. Although we haven't been through L'Hospital rule yet in our lectures, it seems useful. –  Curtain Oct 2 '12 at 17:22

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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Yeah, but is it enough to say that ln(x) is the inverse of $e^x and therefore is growing slower? –  Curtain Oct 2 '12 at 17:26
    
You also have a $x^{5/2}$ inside the $\ln$. How do you know that when you have a greater power inside the $\ln$, then the denominator still increases faster? –  Thomas Oct 2 '12 at 17:28
    
Thomas, good point. I can't possibly know that. –  Curtain Oct 2 '12 at 17:29
    
Polynomial growth dominates logarithmic growth in even the most extreme cases. –  John Roberts Oct 2 '12 at 17:42

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