Using L'hopital's Rule, Find lim_x->infinity (x^2 + 2x + 3)/(3x+2)^2

 Jun 2, 2022

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.



\(\displaystyle \lim_{x\rightarrow\infty }\frac{(x^2 + 2x + 3)}{(3x+2)^2}\\ =\displaystyle \lim_{x\rightarrow\infty }\frac{(2x + 2)}{2(3x+2)(3)}\\ =\displaystyle \lim_{x\rightarrow\infty }\frac{2x + 2}{18x+12}\\ =\displaystyle \lim_{x\rightarrow\infty }\frac{2}{18}\\ =\frac{1}{9} \)

 Jun 3, 2022
edited by Melody  Jun 3, 2022

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