So let's assume $$\frac{2+5\log x}{2+\log x}$$
This function has a vertical asyntote in $x=1/e^2$ but i am able to solve that limit with algebric tricks. So i taught that i can expand the 2 function with Taylor in the centre $x=1/e^2$ and then calculate the limit in order to find out the real limit. My problem is:wich basic function i should expand? Logx or Log(x+1) operating some substitution? Thank's to everybody that will help me.
On
There is no need for Taylor expansion, nor for substitutions.
Since $$ \lim_{x\to e^{-2}}(2+\log x)=0 $$ with $2+\log x>0$ if $x>e^{-2}$, $2+\log x<0$ if $x<e^{-2}$, and $$ \lim_{x\to e^{-2}}(2+5\log x)=-8 $$ we have $$ \lim_{x\to e^{-2}-}\frac{2+5\log x}{2+\log x}=\infty \qquad \lim_{x\to e^{-2}+}\frac{2+5\log x}{2+\log x}=-\infty $$
On
$$L=\lim_{x\to e^{-2}}\frac{2+5\ln(x)}{2+\ln(x)}$$ If we look at the top and bottom individually we see that: $$\lim_{x\to e^{-2}}\left[2+5\ln(x)\right]=2+5(-2)=-8$$ $$\lim_{x\to e^{-2}}\left[2+\ln(x)\right]=0$$ so at first glance it would appear to diverge to $\pm\infty$
if we look a this closer we can rewrite it as: $$\lim_{x\to e^{-2}}\left[\frac{2+5\ln(x)+8}{2+\ln(x)}-\frac{8}{2+\ln(x)}\right]$$ if we use a substitution $y=\ln(x)$ this becomes: $$\lim_{y\to -2}\left[\frac{10+5y}{2+y}\right]-\lim_{y\to -2}\left[\frac{8}{2+y}\right]=5-\lim_{y\to -2}\left[\frac{8}{2+y}\right]$$ and this second limit is clearly divergent to $\pm\infty$ so the limit does not exist
We don't need Taylor indeed by $x=\frac {e^y} {e^2}\to 1$ with $y\to 0$ we have
$$\lim_{x\to \frac 1 {e^2}}\frac{2+5\log x}{2+\log x}=\lim_{y\to 0}\frac{2+5y-10}{2+y-2}=\lim_{y\to 0}\frac{-8+5 y}{ y}=\pm \infty$$
therefore the limit doesn't exist.
Note that by Taylor's expansion at $x=\frac1{e^2}$ we obtain
$$\frac{2+5\log x}{2+\log x}=\frac{-8+5(xe^2-1)+ o(xe^2-1)}{xe^2-1+o(xe^2-1)}=\frac{\frac{-8}{xe^2-1}+5+ o(1)}{1+o(1)}$$
therefore the limit depends upon
$$\lim_{x\to \frac 1 {e^2}} \frac{-8}{xe^2-1}$$
which leads to the same result in a more complicated and convoluted way.