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John45
09-15-2010, 03:42 AM
True or False
When
Lim f(x) exists, it always equals f(a)
x->a

And im having difficulty converting these into expressions, like I feel it's some radical number dicided by another but im not familiar which ones

Sin(1/(2/Pi) = 1
Sin(1/(2/3Pi)
Sin(1/(2/5Pi)
Sin(1/(2/7Pi)
Sin(1/(2/9Pi)
Sin(1/(2/11Pi)

galactus
09-15-2010, 08:17 AM
True or False
When
\lim_{x\to a}f(x) exists, it always equals f(a)

Take \lim_{x\to 0}\frac{sin(x)}{x}. If we just sub in x=0, we get \frac{sin(0)}{0}=\frac{0}{0}.

Yet, this limit exists. It's a rather 'famous' limit used often when dealing with trig limits.

mmm4444bot
09-15-2010, 04:58 PM
True or False

When

Lim f(x) exists, it always equals f(a)
x->a

When we take the limit of function f -- as x approaches the value a -- we never let x actually become a. So, we don't care about the value f(a). Maybe it exists, maybe it doesn't.

We're only interested in whether or not function f approaches some Real constant, as x approaches a.

Clearly, in cases where f(a) does not exist, it cannot possibly be the limit.

And im having difficulty converting these into expressions These are already expressions; the first one is an equation.

What are you trying to do ?

Did they ask you to evaluate these expressions? Is that why you typed the first one equal to 1 ?

(I note also that each sine expression is missing a close parenthesis.)

, like I feel it's some radical number dicided by another Huh ?

What are the instructions that came with these sine expressions ?

Whatever they are, I would expect the first step to be simplification of the inputs.

Sin(1/(2/Pi)) = 1 = sin(Pi/2)

Sin(1/(2/3Pi)) = sin(3Pi/2)

Sin(1/(2/5Pi)) = sin(5Pi/2)

Sin(1/(2/7Pi)) = sin(7Pi/2)

Sin(1/(2/9Pi)) = sin(9Pi/2)

Sin(1/(2/11Pi)) = sin(11Pi/2)

John45
09-15-2010, 09:29 PM
so the answer to the first one is false because it approaches a but doesn't have to be a?

The second one it's asking to make a table of values for sin 1/x

for when x = [2/Pi, 2/3Pi, 2/5Pi, 2/7Pi, 2/9Pi, 2/11Pi]

so I assumed that you'd plug it into your calculator but my answer was incorrect.

akoaysigod
09-15-2010, 09:42 PM
Google unit circle and you'll be able to find the definite values for sin for those angles, and basically all the rest of the angles you'll be dealing with the most. (cosx, sinx)

mmm4444bot
09-15-2010, 10:23 PM
so the answer to the first one is false because it approaches a but doesn't have to be a?

I will upload an image, shortly.

The second one it's asking to make a table of values for sin 1/x (Who knew ? :roll: )

so I assumed that you'd plug it into your calculator Sure, machines are great for generating a table of values.

but my answer was incorrect Obviously, you did something wrong with the calculator.

Just as obvious is the fact that I have not seen what you did, so how may I help you?

mmm4444bot
09-15-2010, 11:05 PM
Double-click image, if needed, to expand.

[attachment=0:24rv1t77]limit.jpg[/attachment:24rv1t77]

The number f(a) does not exist because the number a is not in the domain of this function.

However, the limit of function f (as x practically reaches a) equals the same value as f(a) would, if a were in the domain.

Therefore, in my example, the limit exists, but it does not equal f(a). It is equal to what f(a) would be, were it to exist.

There are many other examples of functions where f(a) does not exist, but the limit does.

mmm4444bot
09-15-2010, 11:15 PM
Oh, hold on.

I understand that you need to make a table of values for sin(1/x) for some common angles x, like Pi/2, 3Pi/2, and 5Pi/2.

Are you telling me that you're putting sin(1/(2/Pi)) into the calculator to get a decimal approximation for sin(1/x) when x = Pi/2 ?

If so, you should be inputting sin(2/Pi), instead !

I mean, if x = Pi/2

then 1/x = 2/Pi, yes?

Subhotosh Khan
09-16-2010, 09:14 AM
True or False
When
Lim f(x) exists, it always equals f(a)
x->a

And im having difficulty converting these into expressions, like I feel it's some radical number dicided by another but im not familiar which ones

Sin(1/(2/Pi) = 1
Sin(1/(2/3Pi)
Sin(1/(2/5Pi)
Sin(1/(2/7Pi)
Sin(1/(2/9Pi)
Sin(1/(2/11Pi)

x = 1/(2/?)

so

1/x = ?/2

sin(1/x) = sin(?/2) = 1

John45
09-20-2010, 11:42 PM
and one last question why does a graphing utility have difficulty plotting the graph of sin (1/x)

these are the choices

A. the function sin (1/x) is undefined near x = 0

B. the alternation between 1 and -1 happens infinitely number of times on the interval (0,h) no matter hwo small h > 0 becomes

C. the function sin (1/x) tends to (positive infinity symbol) near x = 0

i think its b

and does this limit exist and what does it approach?

mmm4444bot
09-21-2010, 12:21 AM
and one last question I see two more questions.

i think its b That's correct, so why are you asking the second question below?

I mean, you're talking about the limit of sin(1/x) as x approaches zero, yes?

You know that a limit only exists when the function approaches some constant, yes?

and does this limit exist?