Question

Can somebody help me with this limit :
(3(lnx)^2 +2x)/(e^x + (1+lnx)^2) when x tends to 0, x>0

Answer 1

Divide both numerator and denominator by (lnx)^2: you get
(9+2x/(lnx)^2)/(e^x/(lnx)^2+1/(lnx)^2+2/lnx+1)
where everything goes to 0 except 9 at the numerator and 1 at the denominator. So you get 9.

Answer 2

\[\lim_{x\rightarrow 0^+}\frac{9+\frac{2x}{(ln(x))^2}}
{

(\frac{e^x}{(ln(x))^2}+\frac{1}{(ln(x))^2}+\frac{2}{ln(x)}+1} = 9
\]
Decided to TeX it out to make it clearer. I agree with @mimmo70. Because\[\lim_{x\rightarrow 0^+} ln(x) = -\infty \] Not certain, but I read the orginal question to not square the 3 in the numerator, so the answer in that case would of course be 3.