# 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!

• @angryavian can you help me with this? – UnusualSkill Nov 30 '14 at 3:45
• @OC-Sansoo can you help me with this? – UnusualSkill Nov 30 '14 at 3:45
• @5xum can help me for this ques? – UnusualSkill Dec 2 '14 at 16:03

Hint

$$\alpha^n-\beta^n=(\alpha-\beta)(\alpha^{n-1}+\beta\alpha^{n-2}+...+\alpha\beta^{n-2}+\beta^{n-1}$$

Set $n=5$, $\alpha=x^{\frac{3}{5}}$ and $\beta=a^{\frac{3}{5}}$ and conclude.

• So finally I set my delta to be (x^12/5+x^1.8a^0.6+x^1.2a^1.2+...)ε/(|x^2+ax+a^2|) is this correct? – UnusualSkill Nov 29 '14 at 12:33

Treat your problem as $$\lim_{x \to a}\sqrt[5]{x^3}=\sqrt[5]{a^3}$$

So you want to simplify

$$|\sqrt[5]{x^3}-\sqrt[5]{a^3}|$$

Treat this as a fraction over $1$, and use the usual trick of "rationalizing the numerator" to get some cancellation. That will make the rest of your proof much easier.

NOTE: As I was typing this, @idm showed the trick that will allow you to rationalize the numerator. Combine what he wrote with what I wrote to do the job.

• But Im not sure how to get rid of the term |x^0.4+x^0.2a^0.2+a^0.4| by estimation,could u help? – UnusualSkill Dec 2 '14 at 11:22
• @UnusualSkill: This is not quite what I recommended. But you can replace your term by its maximum or minimum in the interval $|x-a|<\delta$. Since your expression is monotonic in that interval, substitute $x-\delta$ and $x+\delta$ to get both minimum and maximum (in that order if $a>0$, in reverse order if $a<0$). – Rory Daulton Dec 2 '14 at 12:45
• I don understand what you r talking about can you pls show me?I had been trying to get rid of the term for long time alrdy – UnusualSkill Dec 2 '14 at 13:15

Since $x^{.6} = x^{\frac{6}{10}} = ( x^{1/5})^3$. We can show that

$$\lim_{y \to b} y^3 = b^3 \; \; \; \; \ \ \; \; \; \; \; .......(I)$$

If we establish $(I)$, then putting $y = x^{1/5}$ and $b = a^{1/5}$ gives what you want.

To establish $(I)$, assume we are given some $\epsilon > 0$. Our goal is to find some $\delta > 0$ such that if $|y-b| < \delta$, then $|y^3 - b^3| < \epsilon$. To find such $\delta$, notice

$$|y^3 - b^3| = |y-b| | y^2 +yb + b^2|$$

you can a priori assume $\delta < 1$ and then the tricky part boils down to estimate $|y^2 + yb + b^2|$. But, if we want $|y-b| < \delta$, then we better have

$$|y -b | < 1 \iff b-1 < y < b+1$$

I will let you continue the rest. Once you estimate $|y^2 + yb + b^2|$, say the bound is some $\delta( b)$, just take $\delta = \min \{ 1, \delta(b) \}$.

• Note that you need also to show $\lim_{x\to a} x^{1/5} = a^{1/5}$. – user99914 Nov 30 '14 at 7:19
• @Pedro Arbizu can you tell me how to estimate, I had been trying for a long time.Pls, how to get rid of this term |x^0.4+x^0.2a^0.2+a^0.4| – UnusualSkill Dec 2 '14 at 11:19