# Limit of sin(x<sup>0</sup>)/x as x→0 = π/180

• chandubaba
In summary: The limit would still not exist. I would also point out that you missed the most important value: x= 0.
chandubaba
prove that limit of (sin (x^0)/x) as x tends to zero is π/180(ie pi by 180)

DO pay attention to what you are writing! Or are you comatose, perhaps??

What you've written is utterly meaningless.

If you want to say to a friend: "I really like you", do you say to him "bungafloop-floop"??

That's basically what you've written above in the mathematical language.
Is it really that difficult for you to form a proper sequence of mathematical symbols?

I assume you are talking about the sine function, where the argument is given in degrees, rather than radians, but that does not excuse your cavalier attitude with respect to notation.

by sin x^0/x ,i mean sine x to the power 0 divided by x .i have been using this notation for visual basic.sorry!

Last edited:
Now it's even more confusing than before!
I wasn't aware that visual basic allowed such sloppy, anti-mathematical notation. "sine x to the power 0" is (sin x)^0, not sin x^0 which means "sin of (x^0)= sin(1)". In any case, (sin x)^0= 1 so you are really asking about 1/x as x goes to 0. Of course, that does not converge at all, certainly not to $\pi/180$

If you meant, by x^0, "x written in degrees", as arildno suggested, then, since the value "in radians" would be $\pi x/180$ your sin(x)/x becomes $sin(\pi x/180)/x$ You could multiply both numerator and denominator of that by $\pi /180$ and let $y= \pi x/ 180$ to get
$$\frac{\pi}{180}\frac{sin y}{y}$$
which does have the limit you ask for.

Remember that the degree-circle is NOT a power!
The degree-circle works in a similar manner as "m" for "meters":

2m means "two meters".

chandubaba said:
prove that limit of (sin (x^0)/x) as x tends to zero is π/180(ie pi by 180)
I agree with arild.

Now, there are several ways to "prove" limits, but in this case it is simply easier to use the definition of limits: "As the function of X approaches a singular point from the left and the right, the limit is the point that they are coming close to."

Therefore, we can plot out the point and come to the point it is coming close to. In this case, here's our information:

f(x) = sin(x)/x

We're looking for:

Lim(x->0)[sin(x)/x] = ??

Well, create a chart proof:

From right to left:
x: y:
.1 .998334
.01 .999983
.001 .999999
0 undef
-.001 .999999
-.01 .999983
-.1 .998334

The function as X is approaching 0 is 1. Therefere by the definition of limits:

Lim(x->0)[sin(x)/x] = 1

Q.E.D.

No, it isn't GoldPhoenix.
Most probably, OP is given x as measured in DEGREES, rather than in radians.

I would also point out that the fact that
1 .998334
.01 .999983
.001 .999999
0 undef
-.001 .999999
-.01 .999983
-.1 .998334

looks like it is getting close to 1 proves absolutely nothing about the limit. It would be very easy to make up functions that have exactly those values but are, say, -1000000 for x <b>close</b> to 0 ("close" here meaning "less than 0.00000000001 from 0".

## 1. What is the limit of sin(x0)/x as x→0?

The limit of sin(x0)/x as x→0 is undefined. This is because the denominator becomes 0, which results in an indeterminate form.

## 2. Why is the limit of sin(x0)/x as x→0 equal to π/180?

The limit of sin(x0)/x as x→0 is not equal to π/180. This is because the limit does not exist. However, the value of sin(x0)/x as x approaches 0 is equal to π/180, which is a result of the limit approaching this value from both the positive and negative sides.

## 3. How can the limit of sin(x0)/x as x→0 be calculated?

The limit of sin(x0)/x as x→0 cannot be calculated using traditional limit methods, such as direct substitution or algebraic manipulation. It can only be determined by considering the behavior of the function as x approaches 0 from both the positive and negative sides.

## 4. Is the limit of sin(x0)/x as x→0 equal to 0?

No, the limit of sin(x0)/x as x→0 is not equal to 0. As x approaches 0, the value of sin(x0)/x approaches π/180, not 0. This is because the sine function has a non-zero value at x=0, which affects the overall limit.

## 5. Can the limit of sin(x0)/x as x→0 be evaluated using L'Hopital's rule?

No, L'Hopital's rule is not applicable in this case. This is because the limit of sin(x0)/x as x→0 is not in an indeterminate form that can be rewritten as a fraction of two functions, which is a requirement for using L'Hopital's rule.

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