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