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

ghazi
 one year ago
Best ResponseYou've already chosen the best response.2answer 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

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

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

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

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

ghazi
 one year ago
Best ResponseYou've already chosen the best response.2of 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

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

ghazi
 one year ago
Best 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

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

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

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

ghazi
 one year ago
Best ResponseYou've already chosen the best response.2and 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

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

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

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

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

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

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