# Calculate the limit..

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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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
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

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,

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

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