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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!

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  • $\begingroup$ @mfl can help with this? $\endgroup$ – UnusualSkill Dec 2 '14 at 7:10
  • $\begingroup$ @amWhy can help with this?urgent! $\endgroup$ – UnusualSkill Dec 2 '14 at 10:44
  • $\begingroup$ @Stef can help? $\endgroup$ – UnusualSkill Dec 2 '14 at 10:44
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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 $x\to a$.

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  • $\begingroup$ is there any easier to understand approach?@Greg Martin $\endgroup$ – UnusualSkill Dec 2 '14 at 8:17
  • $\begingroup$ can you finish the proof? I want to see how it is done. really confused $\endgroup$ – UnusualSkill Dec 2 '14 at 9:00
  • $\begingroup$ I do like yours for the next step and estimate the denominator <a^0.8 and my question is how to get rid of |x^0.4+x^0.2a^0.2+a^0.4| by estimation? thx for reply pls $\endgroup$ – UnusualSkill Dec 2 '14 at 10:31

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