Limit of H(x) for between graphs

[furi0n]

Homework Statement

if the graph pf y=h(x) always lies between the graphs of y=x^3 and Y=x^1/3 for what real numbers a can you determine the value of limH(x)x->a? explain and find the limit for each of these values of a

Homework Equations

The Attempt at a Solution

i can draw two graphs but i cannot understand the graph of h(x) there are a lot of real numbers between these two graphs which one should i use

Not: please only help mi finding out values of a .. :)

 

[fzero]

What can you say about h(x) near the point(s) where x³ and x^1/3 approach one another?

 

[furi0n]

excuse me ? does this mean points between [-1,1] interval?

 

[fzero]

excuse me ? does this mean points between [-1,1] interval?

What I mean is that you said you sketched the graphs. Where do the graphs of x³ and x^1/3 meet? Is there anything that you can say about the value of h(x) at those points? Can you say anything about the value of h(x) as $$x\rightarrow \pm \infty$$?

 

[furi0n]

What I mean is that you said you sketched the graphs. Where do the graphs of x³ and x^1/3 meet? Is there anything that you can say about the value of h(x) at those points? Can you say anything about the value of h(x) as $$x\rightarrow \pm \infty$$?

The graphs meet (1,1) and (-1,-1), no there isn't because i don't know anything about h(x). Then İ can't find any limit of numbers, actually ı don't understand what we did. i haven't understand Question yet

 

[HallsofIvy]

Yes, you don't understand the question! If $x^3< h(x)< x^{1/3}$ (which happens for x< -1 and 0< x< 1) then you know that $\lim_{x\to a}x^3< \lim_{x\to a} h(x)\le \lim_{x\to a} x^{1/3}$ which, since $x^3$ and [itex]x^{1/3} are continuous, is the same as $a^3\le \lim_{x\to a} h(x)\le a^{1/3}$.<br /> <br /> If $x^{1/3}< h(x)< x^3$ which happens for -1< x< 0 or x> 1, then, similarly, $a^{1/3}\le h(x)\le a^3$.<br /> <br /> In general, $\lim_{x\to a} h(x)$ could be <b>any</b> number between those bounds. But what happens when $a^3= a^{1/3}$? For what a does that happen?[/itex]

 

[furi0n]

Yes, you don't understand the question! If $x^3< h(x)< x^{1/3}$ (which happens for x< -1 and 0< x< 1) then you know that $\lim_{x\to a}x^3< \lim_{x\to a} h(x)\le \lim_{x\to a} x^{1/3}$ which, since $x^3$ and [itex]x^{1/3} are continuous, is the same as $a^3\le \lim_{x\to a} h(x)\le a^{1/3}$.<br /> <br /> If $x^{1/3}< h(x)< x^3$ which happens for -1< x< 0 or x> 1, then, similarly, $a^{1/3}\le h(x)\le a^3$.<br /> <br /> In general, $\lim_{x\to a} h(x)$ could be <b>any</b> number between those bounds. But what happens when $a^3= a^{1/3}$? For what a does that happen?[/itex]

$<br /> Okey , now i understood then x=1,-1 and 0 limh(x)= 1,-1,0 we learned this theorem today. now i understand this perfectly thanks a lot$