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!

  • $\begingroup$ @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

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 $x\to a$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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