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Dodo1
-infity and infity, whats the difference?
\[\infty -\infty\]\[\frac{ Sqrt(x^2+14x) }{ 14-13x } \]
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
and basically, all the numbers lie between -infinity and +infinity
and i dont know what is the limit of the question that you have given :(
how do you evaluate the limit \[\lim_{x \rightarrow \infty }\sqrt{x^2+3x+1}-x\]
and thank you for the explantion. but is it really matter ( - or + infitiy) when evaluate the limit?
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 infinity-infinity then that is kinda undetermined form
I see, how do i do the first question>?
\[\sqrt{(x^2+3x+1)}-x * \frac{ x+\sqrt{(x^2+3x+1)} }{ x+ \sqrt{x^2+3x+1} }\] simplify this and your answer would be 3/2
what about negeative infity?
the answer is not the same isnt it?
well 3/2 is the answer of \[\lim x> \infty \sqrt{x^2+3x+1}-x\]
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
medels *infinity for you!
thank you very much :D :)
Another way of finding the limit: \[\lim_{x\to-\infty}\left(\sqrt{x^2+3x+1}-x\right)\\ \lim_{x\to-\infty}\left(\sqrt{x^2}\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}-x\right)\\ \lim_{x\to-\infty}\left(|x|\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}-x\right)\\ \text{Since $x\to-\infty, |x|=-x$. If that doesn't make sense,}\\ \text{look up the definition of absolute value.}\\ \lim_{x\to-\infty}\left(-x\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}-x\right)\\ \lim_{x\to-\infty}(-x)\left(\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}+1\right)=-(-\infty)(\sqrt1+1)=\infty\]
\[\lim_{x\to\infty}\frac{\sqrt{x^2+14x}}{14-13x}\\ \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{1+\frac{14}{x}}}{14-13x}\\ \lim_{x\to\infty}\frac{|x|\sqrt{1+\frac{14}{x}}}{14-13x}\\ \text{Since $x\to\infty,|x|=x$.}\\ \lim_{x\to\infty}\frac{x\sqrt{1+\frac{14}{x}}}{14-13x}\\ \lim_{x\to\infty}\frac{\sqrt{1+\frac{14}{x}}}{\frac{14}{x}-13}=\frac{\sqrt1}{-13}=-\frac{1}{13}\]
but why its infy(sqrt1)+1=infy ?
\[-(-\infty)\left(\sqrt1+1\right)=\infty\left(1+1\right)=\infty(2)=\infty\]