MHB Why x^2/(1+x^2) > u Near Infinity: Explained

[evinda]

Hello! :)
Could you explain me why $lim_{x \to \pm \infty} \frac{x^2}{1+x^2}=1$ implies that $\forall u<1, \frac{x^2}{1+x^2}>u$ near to $\pm \infty$ ?

 

[I like Serena]

Hello! :)
Could you explain me why $lim_{x \to \pm \infty} \frac{x^2}{1+x^2}=1$ implies that $\forall u<1, \frac{x^2}{1+x^2}>u$ near to $\pm \infty$ ?

The statement $\lim\limits_{x \to \infty} \frac{x^2}{1+x^2}=1$ is defined to mean that for any $\varepsilon>0$ there is a $N$ such that for any $x>N$:
$$\left|\frac{x^2}{1+x^2} - 1\right| < \varepsilon$$If we pick any $u<1$, we can select $\varepsilon = 1-u$, meaning that there is a $N$ such that for any $x>N$:
$$\left|\frac{x^2}{1+x^2} - 1\right| < 1-u$$
This implies that:
$$1 - \frac{x^2}{1+x^2} < 1-u$$
$$\frac{x^2}{1+x^2} > u$$

 

[evinda]

The statement $\lim\limits_{x \to \infty} \frac{x^2}{1+x^2}=1$ is defined to mean that for any $\varepsilon>0$ there is a $N$ such that for any $x>N$:
$$\left|\frac{x^2}{1+x^2} - 1\right| < \varepsilon$$If we pick any $u<1$, we can select $\varepsilon = 1-u$, meaning that there is a $N$ such that for any $x>N$:
$$\left|\frac{x^2}{1+x^2} - 1\right| < 1-u$$
This implies that:
$$1 - \frac{x^2}{1+x^2} < 1-u$$
$$\frac{x^2}{1+x^2} > u$$

I understand.. :) Thank you very much!

 

[evinda]

The statement $\lim\limits_{x \to \infty} \frac{x^2}{1+x^2}=1$ is defined to mean that for any $\varepsilon>0$ there is a $N$ such that for any $x>N$:
$$\left|\frac{x^2}{1+x^2} - 1\right| < \varepsilon$$If we pick any $u<1$, we can select $\varepsilon = 1-u$, meaning that there is a $N$ such that for any $x>N$:
$$\left|\frac{x^2}{1+x^2} - 1\right| < 1-u$$
This implies that:
$$1 - \frac{x^2}{1+x^2} < 1-u$$
$$\frac{x^2}{1+x^2} > u$$

I am looking again at the exercise at which I found this limit..

It is:

$$f,g: \mathbb{R} \to \mathbb{R}, f(x)=\frac{x^2}{1+x^2}, g(x)=0 \forall x$$

We want to calculate $\displaystyle{ |f-g|_{\mathbb{R}}= \sup \{ \frac{x^2}{1+x^2} , x \in \mathbb{R} \}}$

Why,in order to find this supremum,do we take $\displaystyle{ \lim_{x \to \pm \infty } \frac{x^2}{1+x^2}}$ ? (Thinking) (Thinking)

 

[I like Serena]

We want to calculate $\displaystyle{ |f-g|_{\mathbb{R}}= \sup \{ \frac{x^2}{1+x^2} , x \in \mathbb{R} \}}$

Why,in order to find this supremum,do we take $\displaystyle{ \lim_{x \to \pm \infty } \frac{x^2}{1+x^2}}$ ? (Thinking) (Thinking)

To find the supremum we need to verify all local extrema, all boundary extrema, and all asymptotic extrema.
In this particular case the supremum is reached in an asymptote. (Nerd)

 

[evinda]

To find the supremum we need to verify all local extrema, all boundary extrema, and all asymptotic extrema.
In this particular case the supremum is reached in an asymptote. (Nerd)

Why is it like that? (Thinking)

I thought that we find the monotonicity of $|f-g|$ and if it is increasing we take the maximum of the interval,and if it is decreasing the minimum.. (Sweating)

 

[I like Serena]

Why is it like that? (Thinking)

I thought that we find the monotonicity of $|f-g|$ and if it is increasing we take the maximum of the interval,and if it is decreasing the minimum.. (Sweating)

What is the difference between a maximum and a supremum? (Wondering)

Where do you think the maximum is on the interval $(-\infty, +\infty)$? (Thinking)

 

[evinda]

What is the difference between a maximum and a supremum? (Wondering)

Where do you think the maximum is on the interval $(-\infty, +\infty)$? (Thinking)

The supremum is $+\infty$ and the maximum is an $a \in (-\infty, +\infty)$,right? (Blush)

 

[I like Serena]

The supremum is $+\infty$ and the maximum is an $a \in (-\infty, +\infty)$,right? (Blush)

Erm... no. (Worried)

The maximum is the highest value the function takes for some $a \in (-\infty, +\infty)$.
In this case there is no such value, since there is always a higher value.

The supremum is the lowest upper bound for the maximum.
In this case it is $1$.

 

[evinda]

Erm... no. (Worried)

The maximum is the highest value the function takes for some $a \in (-\infty, +\infty)$.
In this case there is no such value, since there is always a higher value.

The supremum is the lowest upper bound for the maximum.
In this case it is $1$.

I understand! Thank you very much! (Smile)