Question

-infity and infity, whats the difference?

Answer 1

\[\infty -\infty\]\[\frac{ Sqrt(x^2+14x) }{ 14-13x } \]

Answer 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

Answer 3

and basically, all the numbers lie between -infinity and +infinity

Answer 4

and i dont know what is the limit of the question that you have given :(

Answer 5

how do you evaluate the limit
\[\lim_{x \rightarrow \infty }\sqrt{x^2+3x+1}-x\]

Answer 6

and thank you for the explantion.
but is it really matter ( - or + infitiy) when evaluate the limit?

Answer 7

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

Answer 8

I see, how do i do the first question>?

Answer 9

\[\sqrt{(x^2+3x+1)}-x * \frac{ x+\sqrt{(x^2+3x+1)} }{ x+ \sqrt{x^2+3x+1} }\] simplify this and your answer would be 3/2

Answer 10

what about negeative infity?

Answer 11

the answer is not the same isnt it?

Answer 12

well 3/2 is the answer of \[\lim x> \infty \sqrt{x^2+3x+1}-x\]

Answer 13

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

Answer 14

thank you very much!

Answer 15

:D

Answer 16

medels *infinity for you!

Answer 17

thank you very much :D :)

Answer 18

Another way of finding the limit:
\[\lim_{x\to-\infty}\left(\sqrt{x^2+3x+1}-x\right)\\
\lim_{x\to-\infty}\left(\sqrt{x^2}\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}-x\right)\\
\lim_{x\to-\infty}\left(|x|\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}-x\right)\\
\text{Since $x\to-\infty, |x|=-x$. If that doesn't make sense,}\\
\text{look up the definition of absolute value.}\\
\lim_{x\to-\infty}\left(-x\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}-x\right)\\
\lim_{x\to-\infty}(-x)\left(\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}+1\right)=-(-\infty)(\sqrt1+1)=\infty\]

Answer 19

\[\lim_{x\to\infty}\frac{\sqrt{x^2+14x}}{14-13x}\\
\lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{1+\frac{14}{x}}}{14-13x}\\
\lim_{x\to\infty}\frac{|x|\sqrt{1+\frac{14}{x}}}{14-13x}\\
\text{Since $x\to\infty,|x|=x$.}\\
\lim_{x\to\infty}\frac{x\sqrt{1+\frac{14}{x}}}{14-13x}\\
\lim_{x\to\infty}\frac{\sqrt{1+\frac{14}{x}}}{\frac{14}{x}-13}=\frac{\sqrt1}{-13}=-\frac{1}{13}\]

Answer 20

thats really cool too.

Answer 21

but why its infy(sqrt1)+1=infy ?

Answer 22

\[-(-\infty)\left(\sqrt1+1\right)=\infty\left(1+1\right)=\infty(2)=\infty\]