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Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
\[\infty \infty\]\[\frac{ Sqrt(x^2+14x) }{ 1413x } \]
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.2
answer of your first question, what is the difference between  infinity and +infinity, lets say if i ask you to divide a line segment in smaller parts , you will keep chopping it off till it is visible to you but if we asked how smaller we can chop, the answer is infinity and if i ask you to imagine how far you can go in the universe the answer is again infinity, in terms of mathematics when you go in the negative direction of a number line then you tend to approach  infinity and when you go in + direction you approach +infinity
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.2
and basically, all the numbers lie between infinity and +infinity
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.2
and i dont know what is the limit of the question that you have given :(
 one year ago

Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
how do you evaluate the limit \[\lim_{x \rightarrow \infty }\sqrt{x^2+3x+1}x\]
 one year ago

Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
and thank you for the explantion. but is it really matter (  or + infitiy) when evaluate the limit?
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.2
of course it does matter where your function is moving, and if you just see the function that you have given, you can answer it with logic and theoretical explanation that is, when x is tending to infinity the value of your function would be infinite just substitute a huge number and see what you get and if you are getting infinityinfinity then that is kinda undetermined form
 one year ago

Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
I see, how do i do the first question>?
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.2
\[\sqrt{(x^2+3x+1)}x * \frac{ x+\sqrt{(x^2+3x+1)} }{ x+ \sqrt{x^2+3x+1} }\] simplify this and your answer would be 3/2
 one year ago

Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
what about negeative infity?
 one year ago

Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
the answer is not the same isnt it?
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.2
well 3/2 is the answer of \[\lim x> \infty \sqrt{x^2+3x+1}x\]
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.2
and the answer of your first question is infinity because just use L' hospital's rule and see after differentiating you will get 2x/13 now if you're going to substitute infinity at place of x then you will have infinity as answer too
 one year ago

Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
thank you very much!
 one year ago

Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
medels *infinity for you!
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.2
thank you very much :D :)
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
Another way of finding the limit: \[\lim_{x\to\infty}\left(\sqrt{x^2+3x+1}x\right)\\ \lim_{x\to\infty}\left(\sqrt{x^2}\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}x\right)\\ \lim_{x\to\infty}\left(x\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}x\right)\\ \text{Since $x\to\infty, x=x$. If that doesn't make sense,}\\ \text{look up the definition of absolute value.}\\ \lim_{x\to\infty}\left(x\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}x\right)\\ \lim_{x\to\infty}(x)\left(\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}+1\right)=(\infty)(\sqrt1+1)=\infty\]
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
\[\lim_{x\to\infty}\frac{\sqrt{x^2+14x}}{1413x}\\ \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{1+\frac{14}{x}}}{1413x}\\ \lim_{x\to\infty}\frac{x\sqrt{1+\frac{14}{x}}}{1413x}\\ \text{Since $x\to\infty,x=x$.}\\ \lim_{x\to\infty}\frac{x\sqrt{1+\frac{14}{x}}}{1413x}\\ \lim_{x\to\infty}\frac{\sqrt{1+\frac{14}{x}}}{\frac{14}{x}13}=\frac{\sqrt1}{13}=\frac{1}{13}\]
 one year ago

Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
thats really cool too.
 one year ago

Dodo1 Group TitleBest ResponseYou've already chosen the best response.1
but why its infy(sqrt1)+1=infy ?
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
\[(\infty)\left(\sqrt1+1\right)=\infty\left(1+1\right)=\infty(2)=\infty\]
 one year ago
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