+118608 Hi Hipie,
Might I suggest that you change your icon. Members are given priority, by me and by many others. Your icon is so non-descript that it is easy to overlook your membership status.
Using L'hopital's Rule, Find \(\displaystyle \lim_{x\rightarrow\infty }\frac{(x^2 + 2x + 3)}{(3x+2)^2}\)
I will work this properly in a moment but I can see straight off that the answer is goint to be 1/9
I know this because as x tends to infinity only the highest powers of x have any consequence.
So we have
\(\displaystyle \lim_{x\rightarrow\infty }\frac{(x^2 + 2x + 3)}{(3x+2)^2}=\displaystyle \lim_{x\rightarrow\infty }\frac{(x^2 +\dots)}{((3x)^2+\dots) }=\frac{1}{9}\)
ok now I will do it properly.
\(\displaystyle \lim_{x\rightarrow\infty }\frac{(x^2 + 2x + 3)}{(3x+2)^2}\\ =\displaystyle \lim_{x\rightarrow\infty }\frac{(2x + 2)}{2(3x+2)(3)}\\ =\displaystyle \lim_{x\rightarrow\infty }\frac{2x + 2}{18x+12}\\ =\displaystyle \lim_{x\rightarrow\infty }\frac{2}{18}\\ =\frac{1}{9} \)
