How do you solve this limit, $\displaystyle \displaystyle\lim_{x\to{0}}\frac{ln(\sqrt{x^5+3x+1 })}{x}$ without l'Hopital's Rule? Using l'Hopital's rule is too easy.

Printable View

- Dec 3rd 2007, 08:21 AMpolymeraseln Limit
How do you solve this limit, $\displaystyle \displaystyle\lim_{x\to{0}}\frac{ln(\sqrt{x^5+3x+1 })}{x}$ without l'Hopital's Rule? Using l'Hopital's rule is too easy.

- Dec 3rd 2007, 08:59 AMTwistedOne151Re: Limit
First, you can use the properties of logarithms to take $\displaystyle \ln (\sqrt{1+3x+x^5}) = \tfrac{1}{2}\ln (1+3x+x^5)$. Then, you can use the fact that for small x ($\displaystyle |x| \ll 1$), $\displaystyle \ln (1+x) \approx x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$ (the Mercator series), and thus in the limit as x approaches zero, $\displaystyle \ln (1+3x+x^5)$ approaches $\displaystyle 3x-\tfrac{9}{2}x^2+\mathcal{O}(x^3)$, so thus

$\displaystyle \displaystyle\lim_{x\to{0}}\frac{ln(\sqrt{x^5+3x+1 })}{x} = \tfrac{1}{2}\displaystyle\lim_{x\to{0}}\frac{3x-\tfrac{9}{2}x^2+\cdots}{x} = \frac{3}{2}$

Hope this is illuminating.

--Kevin C. - Dec 3rd 2007, 09:26 AMpolymerase
- Dec 3rd 2007, 09:33 AMcolby2152
- Dec 3rd 2007, 09:35 AMTwistedOne151Re: 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 AMpolymerase
- Dec 3rd 2007, 10:07 AMPaulRS
In many cases, like this one, it is enough to consider the following:

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

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

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

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

Since $\displaystyle \frac{\ln(1+3x+5x^2)}{3x+5x^2}\rightarrow{1}$ - Dec 3rd 2007, 11:07 AMpolymerase
- Dec 3rd 2007, 02:47 PMtopsquark
- Dec 3rd 2007, 03:14 PMPaulRS
- Dec 3rd 2007, 04:13 PMgalactus
An interesting observation to make is that, in general,

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

In this case, b/c = 3/2. - Dec 3rd 2007, 05:17 PMThePerfectHacker