I A question about limits and infinity

[0kelvin]

I hope I can make this question clear enough.

When we have a function such as f(x) = 1/x and calculate the side limits at x = 0, the right side goes to positive infinity. The left side goes to negative infinity. In calculus we are pluggin in values closer and closer to zero and seeing what the value of f(x) is. For ex: 1/10, then plug 1/100, then 1/500 so on. Is there a more rigorous way to prove that the function is in fact going to infinity?

another example: f(x) = 1/[sqrt(9 + x) - 3]. If x = 0 we have a division by zero. Now if I plug in something small such as 10^(-10), it's not zero but we are going beyond the precision of a hand calculator. If I plot this graph and zoom in enough, at some point google warns that the graph may be wrong due to precision errors. Is there some theory behind such precision errors and may lead us to think that the graph is increasing or decreasing when it's not?

 

[fresh_42]

I hope I can make this question clear enough.

When we have a function such as f(x) = 1/x and calculate the side limits at x = 0, the right side goes to positive infinity. The left side goes to negative infinity. In calculus we are pluggin in values closer and closer to zero and seeing what the value of f(x) is. For ex: 1/10, then plug 1/100, then 1/500 so on. Is there a more rigorous way to prove that the function is in fact going to infinity?

You show that given any number $M>0$ there will be a value $x>0$ such that $f(x)>M$. Since $M$ was arbitrary large, the function $f(x)$ grows beyond all limits. The negative version is according.

another example: f(x) = 1/[sqrt(9 + x) - 3]. If x = 0 we have a division by zero. Now if I plug in something small such as 10^(-10), it's not zero but we are going beyond the precision of a hand calculator. If I plot this graph and zoom in enough, at some point google warns that the graph may be wrong due to precision errors. Is there some theory behind such precision errors and may lead us to think that the graph is increasing or decreasing when it's not?

I don't understand this. Do you mean error calculations? This case isn't any different from the previous one.

 

[Mark44]

another example: f(x) = 1/[sqrt(9 + x) - 3]. If x = 0 we have a division by zero. Now if I plug in something small such as 10^(-10), it's not zero but we are going beyond the precision of a hand calculator. If I plot this graph and zoom in enough, at some point google warns that the graph may be wrong due to precision errors. Is there some theory behind such precision errors and may lead us to think that the graph is increasing or decreasing when it's not?

Plugging in numbers might not be very helpful, but sketching a graph of $f(x) = \frac 1 {\sqrt{x + 9} - 3}$ would be very helpful.

First, sketch the graph of $y = \sqrt{x + 9} - 3$. This left endpoint of this graph is at (-9, -3) and goes through the origin. This graph is strictly increasing on its domain.

The reciprocal function, $f(x) = \frac 1 {\sqrt{x + 9} - 3}$, will have a vertical asymptote at x = 0. Since the denominator is negative for x < 0, the graph of f goes off to negative infinity. Since the denominator is positive for x > p, the graph of f goes off to positive infinity.

 

[Trollfaz]

You show that given any number $M>0$ there will be a value $x>0$ such that $f(x)>M$. Since $M$ was arbitrary large, the function $f(x)$ grows beyond all limits. The negative version is according.


I don't understand this. Do you mean error calculations? This case isn't any different from the previous one.

You need to use the delta epsilon function basically it means if we add a positive value ,delta ,to x=0 and constrain this value closer to 0 you need to prove that 1/x is will always be larger than M a large constant for all values of delta. From the negative just subtract delta.
x<d ==>1/x>1/d
assign 1/d=M ==>1/x>M

you have to prove that this M exists and can be <= 1/d for all delta

 

[PeroK]

You need to use the delta epsilon function basically it means if we add a positive value ,delta ,to x=0 and constrain this value closer to 0 you need to prove that 1/x is will always be larger than M a large constant for all values of delta. From the negative just subtract delta.
x<d ==>1/x>1/d
assign 1/d=M ==>1/x>M

you have to prove that this M exists and can be <= 1/d for all delta

You mean that finding one value of $x$ is not enough? You need to show that the function is sufficiently large for all $0 < x < \delta$? For some $\delta$ that depends on $M$.

 

[fresh_42]

You need to use the delta epsilon function basically it means if we add a positive value ,delta ,to x=0 and constrain this value closer to 0 you need to prove that 1/x is will always be larger than M a large constant for all values of delta. From the negative just subtract delta.
x<d ==>1/x>1/d
assign 1/d=M ==>1/x>M

you have to prove that this M exists and can be <= 1/d for all delta

The $\varepsilon -\delta$ wording doesn't apply to infinity. The definition of $\longrightarrow \pm \infty$ is different from $\longrightarrow L$ since infinity isn't a number.

 

[PeroK]

The $\varepsilon -\delta$ wording doesn't apply to infinity. The definition of $\longrightarrow \pm \infty$ is different from $\longrightarrow L$ since infinity isn't a number.

That may be so, but it doesn't change the fact that this is not correct:

You show that given any number $M>0$ there will be a value $x>0$ such that $f(x)>M$. Since $M$ was arbitrary large, the function $f(x)$ grows beyond all limits. The negative version is according.

Finding one value of $x > 0$ is not enough. By that definition the function $e^x$ would be unbounded as $x \rightarrow 0$.

 

[fresh_42]

That may be so, but it doesn't change the fact that this is not correct:

Indeed! That was more than just sloppy by me. I totally forgot the neighborhood.