# How to prove the right sided limit of a function

Im having a little trouble understanding how I can solve a question related to proving right sided limits.

I have a function and it's limit as it goes to a given point however I am unsure how to rigorously prove this limit. Here we have right sided limit $$\lim_{x \to \sqrt{2}^+} 2x - x^3 = 0$$ I cannot use the algebra of limits, I would really appreciate if someone could give me a few hints to get me started as I am completely lost on this one.

I hadn't covered $\varepsilon$-$\delta$ proofs by the time I finished calculus. Here is a way (hopefully) to solve this limit without the rigor of $\varepsilon$-$\delta$. We will transform the limit into one that is easier to prove. Let $x = \sqrt{2} + \varepsilon$ where $\varepsilon>0$. Then \begin{align} 2x-x^3 = 2(\sqrt{2} + \varepsilon)-(\sqrt{2} + \varepsilon)^3 \\ =2\sqrt{2}+2\varepsilon-(2\sqrt{2}+4\varepsilon+2\sqrt{2}\varepsilon^2+ \varepsilon^3) \\ = -2\varepsilon-2\sqrt{2}\varepsilon^2-\varepsilon^3 \\ = -\varepsilon(2+2\sqrt{2}\varepsilon +\varepsilon^2) \\ = -\varepsilon(\sqrt{2}+\varepsilon)^2 \end{align} Now notice that $(\lim_{x \to \sqrt{2}^+}x)$ is equivalent to $(\lim_{\varepsilon \to 0^+}\sqrt{2}+\varepsilon)$ and hence $$\lim_{x \to \sqrt{2}^+} 2x-x^3 \equiv \lim_{\varepsilon \to 0^+} -\varepsilon(\sqrt{2}+\varepsilon)^2$$ Can you solve this rewritten limit?
Hint: for all given $\epsilon>0$, try to find a $\delta>0$ such that $$2x-x^3\in (-\epsilon,\epsilon)$$ for all $x\in(\sqrt2,\sqrt2+\delta)$. You can evaluate it without much difficulty.
Definition of right hand side limit: We say that $\lim_{x \rightarrow a^{+}} f(x)=L$, Given $\epsilon >0$ there is $\delta>0$ such that if $a<x<a+\delta$ then $|f(x)-L|<\epsilon$.
You must show that for all $\epsilon>0$, there exists $\delta>0$ such that $$\sqrt{2}<x<\sqrt{2}+\delta \implies |2x-x^3| < \epsilon.$$ You should try to work with the inequality $|2x-x^3| < \epsilon$. For instance, note that $2x-x^3 = x(2-x^2) = x(x-\sqrt{2})(x+\sqrt{2})$. Given that $x-\sqrt{2}$ will be close to 0, you have control over the term $(x-\sqrt{2})$. You may want to try putting a cap on $\delta$ (such as $\delta<1$) so that you can get control over the other two terms.