# ln Limit

• Dec 3rd 2007, 08:21 AM
polymerase
ln Limit
How do you solve this limit, $\displaystyle\lim_{x\to{0}}\frac{ln(\sqrt{x^5+3x+1 })}{x}$ without l'Hopital's Rule? Using l'Hopital's rule is too easy.
• Dec 3rd 2007, 08:59 AM
TwistedOne151
Re: Limit
First, you can use the properties of logarithms to take $\ln (\sqrt{1+3x+x^5}) = \tfrac{1}{2}\ln (1+3x+x^5)$. Then, you can use the fact that for small x ( $|x| \ll 1$), $\ln (1+x) \approx x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$ (the Mercator series), and thus in the limit as x approaches zero, $\ln (1+3x+x^5)$ approaches $3x-\tfrac{9}{2}x^2+\mathcal{O}(x^3)$, so thus
$\displaystyle\lim_{x\to{0}}\frac{ln(\sqrt{x^5+3x+1 })}{x} = \tfrac{1}{2}\displaystyle\lim_{x\to{0}}\frac{3x-\tfrac{9}{2}x^2+\cdots}{x} = \frac{3}{2}$

Hope this is illuminating.

--Kevin C.
• Dec 3rd 2007, 09:26 AM
polymerase
Quote:

Originally Posted by TwistedOne151
First, you can use the properties of logarithms to take $\ln (\sqrt{1+3x+x^5}) = \tfrac{1}{2}\ln (1+3x+x^5)$. Then, you can use the fact that for small x ( $|x| \ll 1$), $\ln (1+x) \approx x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$ (the Mercator series), and thus in the limit as x approaches zero, $\ln (1+3x+x^5)$ approaches $3x-\tfrac{9}{2}x^2+\mathcal{O}(x^3)$, so thus
$\displaystyle\lim_{x\to{0}}\frac{ln(\sqrt{x^5+3x+1 })}{x} = \tfrac{1}{2}\displaystyle\lim_{x\to{0}}\frac{3x-\tfrac{9}{2}x^2+\cdots}{x} = \frac{3}{2}$

Hope this is illuminating.

--Kevin C.

Is there any other way...? I really unfamilar with the series (Mercator, taylor etc.) stuff.
• Dec 3rd 2007, 09:33 AM
colby2152
Quote:

Originally Posted by polymerase
Is there any other way...? I really unfamilar with the series (Mercator, taylor etc.) stuff.

L'Hopital's rule is a nice trick to get past the tough work of using series and other definitions. BTW, why would you not want to use the rule?
• Dec 3rd 2007, 09:35 AM
TwistedOne151
Re: Limit
Not really. The presence of the logarithm ultimately requires either some form of series about the value to which its arguement converges, or else l'Hopital's rule (which essentially derives from the Taylor series for the numerator and denominator for the 0/0 indeterminate form). Ultimately you have to "know" in some fashion the behavior of ln(x) about x=1.
• Dec 3rd 2007, 09:36 AM
polymerase
Quote:

Originally Posted by colby2152
L'Hopital's rule is a nice trick to get past the tough work of using series and other definitions. BTW, why would you not want to use the rule?

If this was simply "find the answer" then of course i used that...but im trying to find "techniques" to imploy for hard limits because most of the time my prof won't let us use l'Hopital...when he does...the function is crazy.
• Dec 3rd 2007, 10:07 AM
PaulRS
In many cases, like this one, it is enough to consider the following:

$\lim_{u\rightarrow{0}}\frac{\ln(u+1)}{u}=\lim_{u\r ightarrow{0}}\ln(1+u)^{u}=1$

Because: $\lim_{u\rightarrow{0}}(1+u)^{u}=e$

Thus: $\lim_{x\rightarrow{0}}\frac{\ln(\sqrt[]{1+3x+5x^2})}{x}=\lim_{x\rightarrow{0}}\frac{\frac {1}{2}\cdot{\ln(1+3x+5x^2)}}{x}=\frac{1}{2}\cdot{\ lim_{x\rightarrow{0}}\frac{\frac{\ln(1+3x+5x^2)}{3 x+5x^2}}{\frac{x}{3x+5x^2}}}$

$\lim_{x\rightarrow{0}}\frac{\ln(\sqrt[]{1+3x+5x^2})}{x}=\frac{1}{2}\cdot{\lim_{x\rightarr ow{0}}\frac{\frac{\ln(1+3x+5x^2)}{3x+5x^2}}{\frac{ x}{3x+5x^2}}}=\frac{1}{2}\cdot{\lim_{x\rightarrow{ 0}}\frac{3x+5x^2}{x}}=\frac{3}{2}$

Since $\frac{\ln(1+3x+5x^2)}{3x+5x^2}\rightarrow{1}$
• Dec 3rd 2007, 11:07 AM
polymerase
Quote:

Originally Posted by PaulRS
In many cases, like this one, it is enough to consider the following:

$\lim_{u\rightarrow{0}}\frac{\ln(u+1)}{u}=\lim_{u\r ightarrow{0}}\ln(1+u)^{u}=1$

Because: $\lim_{u\rightarrow{0}}(1+u)^{u}=e$

Thus: $\lim_{x\rightarrow{0}}\frac{\ln(\sqrt[]{1+3x+5x^2})}{x}=\lim_{x\rightarrow{0}}\frac{\frac {1}{2}\cdot{\ln(1+3x+5x^2)}}{x}=\frac{1}{2}\cdot{\ lim_{x\rightarrow{0}}\frac{\frac{\ln(1+3x+5x^2)}{3 x+5x^2}}{\frac{x}{3x+5x^2}}}$

$\lim_{x\rightarrow{0}}\frac{\ln(\sqrt[]{1+3x+5x^2})}{x}=\frac{1}{2}\cdot{\lim_{x\rightarr ow{0}}\frac{\frac{\ln(1+3x+5x^2)}{3x+5x^2}}{\frac{ x}{3x+5x^2}}}=\frac{1}{2}\cdot{\lim_{x\rightarrow{ 0}}\frac{3x+5x^2}{x}}=\frac{3}{2}$

Since $\frac{\ln(1+3x+5x^2)}{3x+5x^2}\rightarrow{1}$

Nice :D
• Dec 3rd 2007, 02:47 PM
topsquark
Quote:

Originally Posted by polymerase
Nice :D

Just be aware that, in general
$\lim_{x \to a}(f(u(x))) \neq f \left ( \lim_{x \to a}u(x) \right )$

It works for this problem, but it can't be generalized to all functions.

-Dan
• Dec 3rd 2007, 03:14 PM
PaulRS
Quote:

Originally Posted by topsquark
Just be aware that, in general
$\lim_{x \to a}(f(u(x))) \neq f \left ( \lim_{x \to a}u(x) \right )$

It works for this problem, but it can't be generalized to all functions.

-Dan

In this case it is true because of the continuity of the logarithm
• Dec 3rd 2007, 04:13 PM
galactus
An interesting observation to make is that, in general,

$\lim_{x\rightarrow{0}}\frac{ln(ax^{2}+bx+1)}{cx}=\ frac{b}{c}$

In this case, b/c = 3/2.
• Dec 3rd 2007, 05:17 PM
ThePerfectHacker
Quote:

Originally Posted by galactus
An interesting observation to make is that, in general,

$\lim_{x\rightarrow{0}}\frac{ln(ax^{2}+bx+1)}{cx}=\ frac{b}{c}$

In this case, b/c = 3/2.

You should make some restrictions on the coefficients. Like $a\not = 0$ and that sort of stuff.