ln Limit

**polymerase** · December 3rd 2007, 08:21 AM

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.

**TwistedOne151** · December 3rd 2007, 08:59 AM

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.

**polymerase** · December 3rd 2007, 09:26 AM

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.

**colby2152** · December 3rd 2007, 09:33 AM

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?

**TwistedOne151** · December 3rd 2007, 09:35 AM

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.

**polymerase** · December 3rd 2007, 09:36 AM

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.

**PaulRS** · December 3rd 2007, 10:07 AM

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

**polymerase** · December 3rd 2007, 11:07 AM

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

**topsquark** · December 3rd 2007, 02:47 PM

Originally Posted by polymerase

Nice

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

**PaulRS** · December 3rd 2007, 03:14 PM

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

**galactus** · December 3rd 2007, 04:13 PM

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.

**ThePerfectHacker** · December 3rd 2007, 05:17 PM

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.

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