Rate of Growth

• Feb 21st 2010, 06:51 PM
VitaX
Rate of Growth
Determine whether the functions grow at the same rate or if one function grows faster than the other as $\displaystyle x \rightarrow \infty$

$\displaystyle 1)$

$\displaystyle lim_{x\rightarrow \infty} \frac{x\ln x}{x+\ln x}$

$\displaystyle lim_{x\rightarrow \infty} \frac{x\frac{1}{x} + \ln x (1)}{1+\frac{1}{x}} \rightarrow lim_{x\rightarrow \infty}\frac{1+\ln x}{\frac{x+1}{x}} \rightarrow lim_{x\rightarrow \infty} \frac{x +x \ln x}{x+1}$

$\displaystyle lim_{x\rightarrow \infty} \frac{1+ \ln x}{1}$
Since $\displaystyle x + \ln x$ reached 1 first, $\displaystyle x \ln x$ outgrew it. (This correct?)

$\displaystyle 2)$

$\displaystyle lim_{x\rightarrow \infty} \frac{(x^2 + 1)^{\frac{1}{2}}}{x}$

$\displaystyle lim_{x\rightarrow \infty}\frac{\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)}{1} \rightarrow lim_{x\rightarrow \infty} \frac{x(x^2+1)^{-\frac{1}{2}}}{1}$

$\displaystyle (x^2+1)^{\frac{1}{2}}$ outgrows $\displaystyle x$ because $\displaystyle x$ reached 1 first.

Was L' Hopitals rule supposed to be used in both of these? And how exactly do you know when to use the rule and when not to?
• Feb 21st 2010, 10:02 PM
CaptainBlack
Quote:

Originally Posted by VitaX
Determine whether the functions grow at the same rate or if one function grows faster than the other as $\displaystyle x \rightarrow \infty$

$\displaystyle 1)$

$\displaystyle lim_{x\rightarrow \infty} \frac{x\ln x}{x+\ln x}$

$\displaystyle lim_{x\rightarrow \infty} \frac{x\frac{1}{x} + \ln x (1)}{1+\frac{1}{x}} \rightarrow lim_{x\rightarrow \infty}\frac{1+\ln x}{\frac{x+1}{x}} \rightarrow lim_{x\rightarrow \infty} \frac{x +x \ln x}{x+1}$

$\displaystyle lim_{x\rightarrow \infty} \frac{1+ \ln x}{1}$
Since $\displaystyle x + \ln x$ reached 1 first, $\displaystyle x \ln x$ outgrew it. (This correct?)

$\displaystyle 2)$

$\displaystyle lim_{x\rightarrow \infty} \frac{(x^2 + 1)^{\frac{1}{2}}}{x}$

$\displaystyle lim_{x\rightarrow \infty}\frac{\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)}{1} \rightarrow lim_{x\rightarrow \infty} \frac{x(x^2+1)^{-\frac{1}{2}}}{1}$

$\displaystyle (x^2+1)^{\frac{1}{2}}$ outgrows $\displaystyle x$ because $\displaystyle x$ reached 1 first.

Was L' Hopitals rule supposed to be used in both of these? And how exactly do you know when to use the rule and when not to?

Try to improve the clarity of you posts. This one is almost incomprehensible.

Two functions $\displaystyle f(x)$ and $\displaystyle g(x)$ grow at the same rate as $\displaystyle x \to \infty$ if and only if both:

$\displaystyle \lim_{x \to \infty }\frac{f(x)}{g(x)} \ne 0$

$\displaystyle \lim_{x \to \infty }\frac{g(x)}{f(x)} \ne 0$.

Your notes may have a slightly different definition, check it.

(Added in explanation: that the limits are non-zero is not to be taken to imply anything about the existence of the limits other than if they exist they are non-zero)

CB
• Feb 21st 2010, 11:22 PM
VitaX
Quote:

Originally Posted by CaptainBlack
Try to improve the clarity of you posts. This one is almost incomprehensible.

Two functions $\displaystyle f(x)$ and $\displaystyle g(x)$ grow at the same rate as $\displaystyle x \to \infty$ if and if both:

$\displaystyle \lim_{x \to \infty }\frac{f(x)}{g(x)} \ne 0$

$\displaystyle \lim_{x \to \infty }\frac{g(x)}{f(x)} \ne 0$.

Your notes may have a slightly different definition, check it.

CB

Text Notes:

f grows faster than g as $\displaystyle x \rightarrow \infty$ if

$\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)}$

or equivalently, if

$\displaystyle \lim_{x \to \infty} \frac{g(x)}{f(x)}$

We also say that g grows slower than f as $\displaystyle x \rightarrow \infty$

f and g grow at the same rate as $\displaystyle x \rightarrow \infty$ if

$\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)} = L$

where L is finite and positive.

According to this then using L' Hopitals rule on the first problem is correct in order to see which grows faster. For the second one they grow at the same rate. I'll show that now with my new steps:

$\displaystyle \lim_{x \to \infty} \frac{(x^2+1)^{\frac{1}{2}}}{x}$

$\displaystyle \lim_{x \to \infty} \left(\frac{x^2+1}{x^2}\right)^\frac{1}{2}$

$\displaystyle \lim_{x \to \infty} \left(1+\frac{1}{x^2}\right)^\frac{1}{2} = 1$

How does this look now?
• Feb 22nd 2010, 12:39 AM
HallsofIvy
You still haven't said what two functions you are talking about!
• Feb 22nd 2010, 12:52 AM
VitaX
Quote:

Originally Posted by HallsofIvy
You still haven't said what two functions you are talking about!

I thought I said that in the first post when I said which function outgrew the other. My fault then for not stating it properly. Anyways $\displaystyle f(x)=x \ln x$ and $\displaystyle g(x)=x + \ln x$ in the first problem. In the second one $\displaystyle f(x)=\sqrt{x^2+1}$ and $\displaystyle g(x)=x$.
• Feb 22nd 2010, 01:15 AM
CaptainBlack
Quote:

Originally Posted by VitaX
Text Notes:

f grows faster than g as $\displaystyle x \rightarrow \infty$ if

$\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)}$

or equivalently, if

$\displaystyle \lim_{x \to \infty} \frac{g(x)}{f(x)}$

We also say that g grows slower than f as $\displaystyle x \rightarrow \infty$

f and g grow at the same rate as $\displaystyle x \rightarrow \infty$ if

$\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)} = L$

where L is finite and positive.

According to this then using L' Hopitals rule on the first problem is correct in order to see which grows faster. For the second one they grow at the same rate. I'll show that now with my new steps:

$\displaystyle \lim_{x \to \infty} \frac{(x^2+1)^{\frac{1}{2}}}{x}$

$\displaystyle \lim_{x \to \infty} \left(\frac{x^2+1}{x^2}\right)^\frac{1}{2}$

$\displaystyle \lim_{x \to \infty} \left(1+\frac{1}{x^2}\right)^\frac{1}{2} = 1$

How does this look now?

You have missed some vital bits of some of these conditions.

I also think you definitions are deficient, since $\displaystyle f(x)=x (\sin(x)+2)$ and $\displaystyle g(x)=x$ grow at the same rate, but none of the limits exist.

CB
• Feb 22nd 2010, 01:58 AM
VitaX
Quote:

Originally Posted by CaptainBlack
You have missed some vital bits of some of these conditions.

I also think you definitions are deficient, since $\displaystyle f(x)=x (\sin(x)+2)$ and $\displaystyle g(x)=x$ grow at the same rate, but none of the limits exist.

CB

Ok your posts are losing me. I just wrote down what the book and my notes said about rate of growth. And was told to only use L' Hopitals rule when the limits are indeterminate meaning $\displaystyle \frac{\infty}{\infty}$ or $\displaystyle \frac{0}{0}$
Basically I just want to know if I attacked the problems correctly as this chapter is a bit confusing to me when to use L' Hopitals rule and when not to. Did I atleast go about doing the problems correctly?
• Feb 22nd 2010, 03:29 AM
CaptainBlack
Quote:

Originally Posted by VitaX
I thought I said that in the first post when I said which function outgrew the other. My fault then for not stating it properly. Anyways $\displaystyle f(x)=x \ln x$ and $\displaystyle g(x)=x + \ln x$ in the first problem. In the second one $\displaystyle f(x)=\sqrt{x^2+1}$ and $\displaystyle g(x)=x$.

For the first:

$\displaystyle \rho(x)=\frac{f(x)}{g(x)}=\frac{x\ln(x)}{x+\ln(x)} =\frac{\ln(x)}{1+\frac{\ln(x)}{x}}$

Then since the denominator $\displaystyle \to 1$ as $\displaystyle x \to \infty$ we have $\displaystyle \rho(x) \to \infty$ because $\displaystyle \ln(x) \to \infty$ under these conditions. So we conclude that $\displaystyle f(x)$ grows more rapidly than $\displaystyle g(x)$

The second is even easier:

$\displaystyle \rho(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x^2+1}}{x}=\ sqrt{1+\frac{1}{x^2}}$

CB
• Feb 22nd 2010, 01:50 PM
VitaX
Quote:

Originally Posted by CaptainBlack
For the first:

$\displaystyle \rho(x)=\frac{f(x)}{g(x)}=\frac{x\ln(x)}{x+\ln(x)} =\frac{\ln(x)}{1+\frac{\ln(x)}{x}}$

Then since the denominator $\displaystyle \to 1$ as $\displaystyle x \to \infty$ we have $\displaystyle \rho(x) \to \infty$ because $\displaystyle \ln(x) \to \infty$ under these conditions. So we conclude that $\displaystyle f(x)$ grows more rapidly than $\displaystyle g(x)$

The second is even easier:

$\displaystyle \rho(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x^2+1}}{x}=\ sqrt{1+\frac{1}{x^2}}$

CB

So I did manage to do them both in that way, one last question would be how can you conclude what the limit is of the second portion? I got 1 after looking at some examples in the book though I'm not very positive in my answer. Unless you just say as x goes to infinity that it goes to 0 and what you are left over with is just sqrt of 1.
• Feb 22nd 2010, 09:14 PM
CaptainBlack
Quote:

Originally Posted by VitaX
So I did manage to do them both in that way, one last question would be how can you conclude what the limit is of the second portion? I got 1 after looking at some examples in the book though I'm not very positive in my answer. Unless you just say as x goes to infinity that it goes to 0 and what you are left over with is just sqrt of 1.

As $\displaystyle 1/x^2$ goes to zero as $\displaystyle x$ goes to infinity $\displaystyle \sqrt{1+1/x^2}$ goes to $\displaystyle 1$ is what I would normally put.

CB