# Prove $\lim_{x \to a}x^{0.6}=a^{0.6}$ using the calculus epsilon-delta definition of limit@@

Let $a>0$. Prove

$$\lim_{x \to a}x^{0.6}=a^{0.6}$$

What I have done:

$$|x^{0.6}-a^{0.6}|=|x^{0.2}-a^{0.2}| \cdot |x^{0.4}+x^{0.2}a^{0.2}+a^{0.4}|$$

Then I am not sure how to continue, I don't know how to get rid of the complicated terms on the RHS

Anyone can help? appreciate!

• @mfl can help with this? – UnusualSkill Dec 2 '14 at 7:10
• @amWhy can help with this?urgent! – UnusualSkill Dec 2 '14 at 10:44
• @Stef can help? – UnusualSkill Dec 2 '14 at 10:44

If you'd like to continue this approach, you can write \begin{align*} |x^{0.6}-a^{0.6}| &= |x^{0.2}-a^{0.2}| \cdot |x^{0.4}+x^{0.2}a^{0.2}+a^{0.4}| \\ &= \frac{|x-a|}{|x^{0.8}+x^{0.6}a^{0.2}+x^{0.4}a^{0.4}+x^{0.2}a^{0.6}+a^{0.8}|} \cdot |x^{0.4}+x^{0.2}a^{0.2}+a^{0.4}|. \end{align*} The function $|x^{0.4}+x^{0.2}a^{0.2}+a^{0.4}|\big/|x^{0.8}+x^{0.6}a^{0.2}+x^{0.4}a^{0.4}+x^{0.2}a^{0.6}+a^{0.8}|$ is bounded for $x$ near $a$, and so by the squeeze theorem, the whole expression tends to $0$ as $x\to a$.