0

So i have a question regarding substitution. It is quite obvious I have to do the substitution $u=1/x$ to solve this limit. However, I do not understand one thing. Substitution is valid because of the theorem mentioned here Limits and substitution. Clearly if we use that theorem $\lim 1/x$ where $ x \to \infty$ is equal to $0$ not $0^+$. Therefore, by the theorem $\lim_{x \to \infty} x^2sin(1/x)=\lim_{u \to 0} sin(u)/u^2$ Where this limit does not exist since depending on which side we approach it we get negative or positive infinity. I know that for some reason we say the limit of 1/x is $0^+$ but i am not sure why. Any explanation would be appreciated.

Thank you.

JohnColtraneisJC
  • 1,919
Sorfosh
  • 3,266
  • @Dr.SonnhardGraubner I know it is, but i don't know why we are allow to say $u \to 0^+$ instead of $u \to 0$ – Sorfosh Jan 17 '18 at 17:48

2 Answers2

1

If you have $x\to +\infty$ and you take $u=\tfrac{1}{x}$, then indeed $u\to 0$ but note that $u>0$...!

Perhaps it's more clear if you consider it the other way around: with $x=\tfrac{1}{u}$, where do you need $u$ to tend to in order for $x \to +\infty$? That's not simply $0$ because for $u \to 0^-$, you would have $x=\tfrac{1}{u}\to -\infty$. So if you want $x=\tfrac{1}{u}\to +\infty$, then you need $u\to 0^+$, the right-hand limit.

StackTD
  • 27,903
  • 34
  • 63
  • $x=1/u$ is not longer the same substitution. I am still confused sorry. $u$ tends to zero, and suddenly i am cherry picking $0^+$ instead of $0$ or $0^-$. – Sorfosh Jan 17 '18 at 16:59
  • Don't get me wrong. I can see why we chose $0^+$ since we are approxhig 0 from the positives. But the theorem I mention does not mention from which way we are approaching the limit but what is the limit of the substitution – Sorfosh Jan 17 '18 at 17:08
0

You want to study if possible, the limit of $\displaystyle x \mapsto x^2\sin\left(\frac{1}{x}\right)$ but where ? $+ \infty$ ? $-\infty$ ?

We'll do both.

$\bullet$ First, if $ \ x \rightarrow +\infty$, with the substitution $\displaystyle u=\frac{1}{x}$ you have $\displaystyle u \rightarrow 0^{+}$. Then $$ \frac{\sin\left(u\right)}{u^2} \underset{u \rightarrow 0^{+}}{\rightarrow}+\infty $$

$\bullet$ Now if $ \ x \rightarrow -\infty$ the same substitution leads to $$ \frac{\sin\left(u\right)}{u^2} \underset{u \rightarrow 0^{-}}{\rightarrow}-\infty $$ You could have noticed that the function $\displaystyle x \mapsto x^2\sin\left(\frac{1}{x}\right)$ is odd. Hence if it diverges to $+\infty$ when $x \rightarrow +\infty$ then it diverges to $-\infty$ when $x \rightarrow -\infty$. Then the result is not a surprise.

See the graph below to convince yourself enter image description here

Atmos
  • 7,369
  • that is true I know I am approaching from 0+ the problem is the theorem that allows substitution claims I should write it goes to 0 – Sorfosh Jan 17 '18 at 17:24
  • $0$ because it is the same if you approach from the right or from the left. – Atmos Jan 17 '18 at 17:29
  • But then the limit I mention in the post does not exist. Since we get negative and positive infinity depending on how we approach it. – Sorfosh Jan 17 '18 at 17:34
  • You are right i've edited. – Atmos Jan 17 '18 at 17:40
  • However, the limit I mention in the post does exist! And it is positive infinity. Therefore, for some reason, we can say $u$ tends to $0^+$ instead of $0$. I am sorry if i am being dense – Sorfosh Jan 17 '18 at 17:48
  • I've edited, hope it will help you understand. In fact, you cannot really right that $x \rightarrow \infty$. It always depends on whether it is $\pm \infty$. – Atmos Jan 17 '18 at 18:00
  • I am sorry, I think you misunderstood me. WhAt I am confused about is exactly why can we say u tends to 0^+. It tends to 0z I am looking for a rigid proof as to why that is a legal move. The theorem I know says I have to say I tends to 0. – Sorfosh Jan 17 '18 at 18:02