Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For example to prove $\lim_{x\to0}f(x) = \lim_{x\to0}f(x^{3})$

Assume $\lim_{x\to0}f(x) = l$ that is $\forall \epsilon > 0$ $\exists \delta > 0$ such that $\forall x$: $0 < |x| < \delta_{1} \implies |f(x)-l|$

Then to show $\lim_{x\to0}f(x^{3}) = l$ which is by replacing $x$ with $x^{3}$ in the assumption: that is $0 < |x| < |x^{3}| < \delta_{2} \implies |f(x^{3})-l|$ for $x< 0$ and $x>1$ and

$0 < |x^{3}| < |x| < 1 \implies |f(x^{3})-l|$ for $0 < x < 1$

that is choosing $\delta_{2} = min(1, \delta_{1})$

thus showing: $\lim_{x\to0}f(x^{3}) = l$

Why would this argument not work if an attempt was made to prove $\lim_{x\to0}f(x) = \lim_{x\to0}f(x^{2})$ which by my question here: Give an example about limits: obviously would not work. Or is there something I overlooked in this proof?

Thanks.

share|improve this question

1 Answer 1

up vote 0 down vote accepted

$x^2$ isn't invertible around $0$, while $x^3$ is. That matters if the left-sided limit of $f(x)$ doesn't exist, while the right-sided one does. In that case, $\lim_{x\rightarrow 0}f(x^2)$ may exist even if $\lim_{x\rightarrow 0}f(x)$ doesn't. One could conjecture though that if both limits exists, they must be equal.

share|improve this answer
1  
That conjecture would be true as $\displaystyle\lim_{x\to 0}f(x^2) = \lim_{x\to 0^+}f(x) = \lim_{x\to 0}f(x)$ under the assumption that the last limit exists. –  Michael Albanese Oct 11 '12 at 13:05
    
So assuming both limit exists: this argument can be used to prove $\lim_{x\to0}f(x) = \lim_{x\to0}f(x^{n})$ for all n? –  mathnoob Oct 11 '12 at 13:46
    
@mathnoob Yes. If you know that $\lim_{x\rightarrow 0}f(x)$ exists, then $\lim_{x\rightarrow 0}f(x^n)$ exists for all $n$ and is equal to $\lim_{x\rightarrow 0}f(x)$ . –  fgp Oct 11 '12 at 13:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.