Limit Help: sqr (9-y^2) / ( y + 3 ) as y approaches -3

[Marco]

I need to find the limit, as y approaches -3, of this expression
sqr (9-y^2) / ( y + 3 )

here's what I did :

[sqr (9-y^2) / ( y + 3 )] / [sqr (9-y^2) / sqr (9-y^2)] = (9-y^2)/ [(y+3) (sqr(9-y^2)] = (3-y) / sqr(9-y^2) =

limit (3-y)/ sqr(9-y^2) , I will subtitute the -3 to the y : [3-(-3)] / sqr (9-(-3)^2) = 6/ sqr (9+9) =6/9 = 2/3
y-> -3

so, the limit of sqr (9-y^2) / ( y + 3 ) as y approaches -3 is 2/3


or


Limit (3-y)/ sqr(9-y^2) , I will subtitute the -3 to the y : [3-(-3)] / sqr (9-(-3)^2) = 6 / sqr (9-(9) = 6/0 = ∞
y -> -3

the limit of sqr (9-y^2) / ( y + 3 ) as y approaches -3 is ∞



which is which?

thank you for replying, thank you very much

 

[mmm4444bot]

… [3-(-3)] / sqr (9-(-3)^2) = 6/ sqr (9+9) = 6/9

Substituting y = -3 in your work leads to 6/0, not 2/3.

I see two issues.

1) The red expression does not equal the blue expression.

2) sqr(9+9) does not equal 9.

---

Hint: Consider the domain of y, first. Then, remember (from definition) that this limit exists only if the same value is achieved as y approaches -3 from each direction. :cool:

 

[mmm4444bot]

I need to find the limit, as y approaches -3, of this expression
sqr (9-y^2) / ( y + 3 )

here's what I did :

[sqr (9-y^2) / ( y + 3 )] / [sqr (9-y^2) / sqr (9-y^2)] = (9-y^2)/ [(y+3) (sqr(9-y^2)] = (3-y) / sqr(9-y^2) =

limit (3-y)/ sqr(9-y^2) , I will subtitute the -3 to the y : [3-(-3)] / sqr (9-(-3)^2) = 6/ sqr (9+9) =6/9 = 2/3
y-> -3

so, the limit of sqr (9-y^2) / ( y + 3 ) as y approaches -3 is 2/3


or


Limit (3-y)/ sqr(9-y^2) , I will subtitute the -3 to the y : [3-(-3)] / sqr (9-(-3)^2) = 6 / sqr (9-(9) = 6/0 = ∞
y -> -3

the limit of sqr (9-y^2) / ( y + 3 ) as y approaches -3 is ∞

In my first reply, I showed where your evaluations contain arithmetic errors.

While I was typing, you edited your original post, by adding the violet part.

which is which?

The answer is neither 2/3 nor ∞.

By the way, 6/0 does not equal ∞.

Can you state the domain of y? You can use the domain to answer this exercise, without any further calculations.

 

[Marco]

In my first reply, I showed where your evaluations contain arithmetic errors.

While I was typing, you edited your original post, by adding the violet part.


The answer is neither 2/3 nor ∞.

By the way, 6/0 does not equal ∞.

Can you state the domain of y? You can use the domain to answer this exercise, without any further calculations.

Thank you for responding sir , Sir, what do you mean the domain of y? I don't get it sorry sir,

My answer is 6/0 or ∞ but they are all wrong, does my steps are wrong sir?

thank you again for responding sir!

 

[HallsofIvy]

I need to find the limit, as y approaches -3, of this expression
sqr (9-y^2) / ( y + 3 )

here's what I did :

[sqr (9-y^2) / ( y + 3 )] / [sqr (9-y^2) / sqr (9-y^2)] = (9-y^2)/ [(y+3) (sqr(9-y^2)] = (3-y) / sqr(9-y^2) =

limit (3-y)/ sqr(9-y^2) , I will subtitute the -3 to the y : [3-(-3)] / sqr (9-(-3)^2) = 6/ sqr (9+9) =6/9 = 2/3
y-> -3

so, the limit of sqr (9-y^2) / ( y + 3 ) as y approaches -3 is 2/3

9- (-3)^2= 9- 9= 0, not 9. Even if that were right, sqrt(9+ 9)= sqrt(2(9))= 3 sqrt(2), not 9.

or


Limit (3-y)/ sqr(9-y^2) , I will subtitute the -3 to the y : [3-(-3)] / sqr (9-(-3)^2) = 6 / sqr (9-(9) = 6/0 = ∞
y -> -3

the limit of sqr (9-y^2) / ( y + 3 ) as y approaches -3 is ∞

I would say, instead, that "the limit does not exist.

which is which?

thank you for replying, thank you very much

Although I would not phrase it that way, the second is correct, the first is not.

 

[Marco]

9- (-3)^2= 9- 9= 0, not 9. Even if that were right, sqrt(9+ 9)= sqrt(2(9))= 3 sqrt(2), not 9.


I would say, instead, that "the limit does not exist.



Although I would not phrase it that way, the second is correct, the first is not.

Thank you very much for responding sir

 

[mmm4444bot]

… what do you mean the domain of y?

The domain of a function is the set of all possible values of the independent variable for which the function is defined.

In your exercise, the independent variable is y. The given function is not defined for all values of y because we cannot take the square root of a negative number.

If you solve the inequality 9 - y^2 ≥ 0, you will discover that your function is not defined when y is less than -3. We cannot talk about a limit, as y approaches -3 from below, if the function doesn't even exist there!

Do you know the difference between a limit and a one-sided limit? Your exercise asks for a limit (i.e., a double-sided limit); it does not ask for a one-sided limit. Therefore, in order for a (double-sided) limit to exist, as y approaches -3, the function must exist on both sides of -3.

Do you know that every limit is a fixed number? Infinity is not a number; therefore, no limit can equal infinity.

When mathematicians write something like limit=∞, they are not saying that the limit is infinity. They are being LAZY. The statement limit=∞ is an abbreviation for: "The limit does not exist because the function's value increases without bound."

The only answer to your exercise is: The limit does not exist.

It does not exist because the function itself does not exist, when y is less than -3. :cool:

 

[mmm4444bot]

Although I would not phrase it that way, the second [limit=∞] is correct …

Halls, were you thinking the exercise asked for a one-sided limit, as y approaches -3 from the right?

 

[Marco]

The domain of a function is the set of all possible values of the independent variable for which the function is defined.

In your exercise, the independent variable is y. The given function is not defined for all values of y because we cannot take the square root of a negative number.

If you solve the inequality 9 - y^2 ≥ 0, you will discover that your function is not defined when y is less than -3. We cannot talk about a limit, as y approaches -3 from below, if the function doesn't even exist there!

Do you know the difference between a limit and a one-sided limit? Your exercise asks for a limit (i.e., a double-sided limit); it does not ask for a one-sided limit. Therefore, in order for a (double-sided) limit to exist, as y approaches -3, the function must exist on both sides of -3.

Do you know that every limit is a fixed number? Infinity is not a number; therefore, no limit can equal infinity.

When mathematicians write something like limit=∞, they are not saying that the limit is infinity. They are being LAZY. The statement limit=∞ is an abbreviation for: "The limit does not exist because the function's value increases without bound."

The only answer to your exercise is: The limit does not exist.

It does not exist because the function itself does not exist, when y is less than -3. :cool:

I see, thank you very much for your response sir it helps me a lot. Thank you very much

 

[mmm4444bot]

Thank you very much

You're welcome.

I noticed that I had misstated something, so I edited my wording.

I had said that, in order for any limit to exist, as y approaches -3, the function must exist on each side of -3. I should not have said "any" limit because I meant "any double-sided" limit.

\(\displaystyle f(y) = \dfrac{\sqrt{9 - y^{2}}}{y + 3}\)


\(\displaystyle \displaystyle\lim_{y \, \to \, -3^-} f(y) = \text{DNE}\)

This notation describes a one-sided limit. The superscript - indicates that y approaches -3 from below. (This limit Does Not Exist because those values of y are not in the domain's function.)


\(\displaystyle \displaystyle\lim_{y \, \to \, -3^+} f(y) = \infty \;\; \text{(DNE)}\)

This notation describes a one-sided limit. The superscript + indicates that y approaches -3 from above. (This limit Does Not Exist because f(y) does not approach a fixed value.)


\(\displaystyle \displaystyle\lim_{y \, \to \, -3} f(y) = \text{DNE}\)

This notation describes a limit. The lack of a superscript - or + indicates that y approaches -3 from each direction. (This limit Does Not Exist because f(y) does not approach the same fixed value in each direction.)

You can google phrases "one-sided limit" and "double-sided limit" to read more about it.