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Make a rough sketch of the curve y = x^n (n is an integer) for the following five...

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Make a rough sketch of the curve y = x^n (n is an integer) for the following five cases:

(i) n = 0

(ii) n > 0, n odd

(iv) n < 0, n odd

Then use these sketches to find the following limits. 

(i) lim as x - infinity of x^n

(ii) lim as x - left side of 0 of x^n

(iv) lim as x - negative infinity of x^


I am not really sure how to approach this. How do I know what x^(any power looks like on a graph)??? Thanks for your time.

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oldnick's profile pic

Posted (Answer #1)

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1)  n=0 the function became: `y=1` straight line  parallel X axis.

    So in this case `lim_(x->oo) x^n=1`


         `lim_(x->0^+)x^n=1```  `lim_(x->-oo)x^n=1`

2)  n>0 odd,  the odd exponential functions are one to one.


     as  `y=x^3` in the graph se can see  that:


       `lim_(x->0^+)x^3=0`    `lim_(x->-oo) x^3=-oo`

as any other odd exponential with n>0

The fuction x^5 does too:

`lim_(x->oo)x^5=oo` `lim_(x->0^+) x^5=0`  `lim_(x->-oo)x^5=-oo`

So we can claim without fall:

`lim_(x->oo)x^n=oo`   `lim_(x->0^+)x^n=0`    `lim_(x->-00)x^n= -oo`

3) the even exponent we use as example easer out of them:`y=x^2`

The bahaviour of  this kind of funcions looks like odd function, but the value `x->-oo`  where we can see funcion tends to `oo` instead  `-oo`

Function `y= x^4` looks like confirm what  previous function suggests.

So we get:  

`lim_(x->oo)x^n=oo`   `lim_(x->0^+) x^n=0`    `lim_(x->-oo)x^n=oo`

4) n odd less than 0.   We know that  `x^(-n)= (1/x)^n`

So the first problem we may get is the value for x =0

I.e   `y= x^(-3)`

It show an hyperbolic function that tend to lay on x axis when x runs trough  `oo`   or `-oo` So that:

`lim_(x->oo)x^n=0`   `lim_(x->0^+) x^n= -oo`   `lim_(x->-oo)x^n=0`

5)Last case is:  `y=x^n`  wit n<o and even.

even in this case we can write  `x^-n=(1/x)^n` , but keeps the  symmetry as the even function with  n>0

It behavour the same of  a negative integer exponent ecept the symmetry respect Y axis. So that:

`lim_(x->oo)x^n=0`   `lim_(x->0^+)x^n= oo`  `lim_(x->-oo)x^n=0` 



pramodpandey's profile pic

Posted (Answer #2)

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(i) if n=0 then

`y=x^n`  ,will be 

y=1   , which a straight line parallel to x axis ,one unit positive above origin.This will pass through first and second quadrant.

(ii)If   n >0 ,and positive integer

`y=x^n`  then graph depends on   x

x=0    then y=0 i.e origin.

if  `x in (-1,0)`

then  y converges to 0 as `n->oo`

if `x in(-oo,-1]`

then y diverges to `-oo` as `n->oo`

If `x in(0,1) `

then y converges to 0  as `n->oo`

if `x in[1,oo)`

then y diverges to `oo`  as `n->oo`

(iii) if n<0 and integer


`y=1/x^n` ,

If x=0 then y is not defined  graph has singularty at x=0

if `x in(0, 1)`  then as `n->oo`  ,`y=1/x^n`  diverges to `oo`

if `x in(1,oo)` then as `n->oo ` ,`y=1/x^n`   converges to 0.

if `x in (-1,0)`  then as  `n->oo` , `y=1/x^n ` 

if `x in(-oo,-1)` then as `n->oo`  ,`y=1/x^n`  converges to 0.

we can discuss same way for x=-1,+1 either converge to -1 or +1.

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