Prove $\lim_{x\to 0^+}{\frac{x^3}{|x|}} = 0$

My work:

Because we are approaching $0$ from the right side, we can drop the absolute value since $x > 0$ $\forall x \in (0,+\infty)$ so we get:

$f(x) = x^2$.

Obviously, if we just plugged in $0$ we would see the $\lim_{x\to 0^+} {x^2} = 0$ but I want to prove this using the following definition:

Let $f$ be a function defined on a subset $S$ of $\mathbb{R}$, let $a$ be a real number that is the limit of some sequence in $S$, and let $L$ be a real number. then $\lim_{x\to a^S}{f(x)} = L$ if and only if for each $\epsilon > 0$ there exists $\delta >0$ such that $x \in S$ and $|x-a| < \delta$ imply $|f(x) - L| < \epsilon$

Using this definition, we find that in order for this limit to exist, we must have that $x_0 \in S$ and $|x-0| < \delta \implies \left|\frac{x^3}{|x|} - L\right| < \epsilon$

However, I have no idea how to continue!

  • $\begingroup$ Since you're using a one sided limit, it should be $\forall \epsilon > 0, \exists \delta > 0, \forall x \in S (x \in (a,a+\delta) \rightarrow |f(x) - L| < \epsilon)$ $\endgroup$ – MathNewbie Jun 5 '15 at 22:19
  • $\begingroup$ I'm a little confused by "$x \in S(x \in (a,a+\delta)$ $\endgroup$ – David South Jun 5 '15 at 22:20
  • 1
    $\begingroup$ Here's another way to read what I wrote above. For each $\epsilon > 0$ there's a positive number $\delta$ so that if $x \in (a,a+\delta)$, then you have $|f(x)-L| < \epsilon$. Using this for the problem you have at hand. Suppose $\epsilon$ is a positive number. You just need to find a positive number $\delta$ so that if $x \in (0,\delta)$, you have $|\frac{x^3}{|x|} - 0| < \epsilon$. This last equation simplifies to $|x^2| < \epsilon$ due to the definition of absolute value. $\endgroup$ – MathNewbie Jun 5 '15 at 22:23
  • $\begingroup$ Ok, it's just a slight twist on the definition I gave above then. Thank you! This was very helpful. $\endgroup$ – David South Jun 5 '15 at 22:26
  • $\begingroup$ the definition is fine for this problem. It's used for two sided limits. But as far as precision goes the one I gave would be something used to quantify one sided limits and it just follows from modifying the $0 < |x-a| < \delta$ part. $\endgroup$ – MathNewbie Jun 5 '15 at 22:27

Some number $\epsilon>0$ is given. You want $|\frac{x^3}{|x|}-L| = |x^2-0| = x^2<\epsilon$. This is the same as wanting $x<\sqrt{\epsilon}$.

So what if you choose $\delta=\sqrt{\epsilon}$?

  • $\begingroup$ Wow, I totally forgot $L$ is the value of the limit. Thank you! But, could you explain what you mean by "Some number $\epsilon > 0$ is given"? What do you mean it is given? $\endgroup$ – David South Jun 5 '15 at 22:22
  • $\begingroup$ @David, it's just another way to say "Let $\epsilon > 0$". $\endgroup$ – MathNewbie Jun 5 '15 at 22:33
  • $\begingroup$ @DavidSouth, yeah, I mean that you have some $\epsilon>0$, and that your job is to choose a $\delta>0$, such that the definition of limits hold. $\endgroup$ – Mankind Jun 5 '15 at 22:34

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