## Homework Statement

Limit sinx^0/x = ?
x-> 0

## Homework Equations

i think this might be useful in evaluating that limit x-> 0 sinx/x = 1

## The Attempt at a Solution

Sorry couldn't figure out how to use the sinx/x limit here,as i tried to use allied angles but couldn't get off that x^0...also i want to know,as power 0 is always = 1,can we directly write 1 instead of x^0? i wrote that but couldn't figure out what to do next?

Mark44
Mentor

## Homework Statement

Limit sinx^0/x = ?
x-> 0

## Homework Equations

i think this might be useful in evaluating that limit x-> 0 sinx/x = 1
I'm confused. Are you trying to evaluate ##\lim_{x \to 0} \frac{sin(x^0)}x ## or ## \lim_{x \to 0} \frac{sin(x)}x##?

If you're trying to evaluate the latter limit, the first limit would not be helpful and is actually a different limit.

## The Attempt at a Solution

Sorry couldn't figure out how to use the sinx/x limit here,as i tried to use allied angles but couldn't get off that x^0...also i want to know,as power 0 is always = 1,can we directly write 1 instead of x^0? i wrote that but couldn't figure out what to do next?

I'm confused. Are you trying to evaluate ##\lim_{x \to 0} \frac{sin(x^0)}x ## or ## \lim_{x \to 0} \frac{sin(x)}x##?

If you're trying to evaluate the latter limit, the first limit would not be helpful and is actually a different limit.

sir I said that I didn't quite get the idea how to use the fundamental radian limit theorem sinx/x as x approaches 0...the LIMIT which I actually want to evaluate is the former one that you stated

You stated you know that x^0 is 1 (For any number x! it is always 1). So now you can write [sin(1)]/x and simply evaluate the limit. What happens when a number in the denominator gets very small as in this limit?

STEMucator
Homework Helper
Remember to check limits from BOTH sides in this case since really what you want is :

##sin(1) lim_{x→0} \frac{1}{x}##

Mark44
Mentor
You stated you know that x^0 is 1 (For any number x! it is always 1).
00 is an indeterminate form.
The one sided limit ##\lim_{x \to 0^+}x^0 = 1##, but you run into problems on the left-sided limit. (See the section titled "The form 00" in http://en.wikipedia.org/wiki/Indeterminate_form.
So now you can write [sin(1)]/x and simply evaluate the limit. What happens when a number in the denominator gets very small as in this limit?

1 person
I think that OP meant x degrees (x°) but clarification is required. If it is x degrees, it is easy to evaluate.

mfb
Mentor
I think that OP meant x degrees (x°) but clarification is required. If it is x degrees, it is easy to evaluate.
Not easier than the regular limit of sin(x)/x, I think. And it does not fit to the explanations in the first post.

kashan123999 said:
also i want to know,as power 0 is always = 1,can we directly write 1 instead of x^0?
We can. For the limit, you only care about x-values close to, but not at the limit (here: 0). Therefore, you don't have to care about 00.

HallsofIvy
Homework Helper
The limit $\lim_{x\to 0}\frac{sin(x)}{x}= 1$ is only valid for x in radians so if kashan123999's "$x^o$" really does mean that x is measured in degees, then we need to convert to radians. $\pi$ radians correspond to 180 degrees so "x" degrees is the same as $x\pi/180$ radians. That is $sin(x^o)$ would be equivalent to $sin(x\pi/180)$. Now, the "x" outside the trig function is a number, not a measurement so is not in "radians" or "degrees". We have, so far, $\lim_{x\to 0}\frac{sin(x\pi/180)}{x}$.

Now, multiply and divide by $\pi/180$ to change the limit to $\frac{\pi}{180}\lim_{x\to 0}\frac{sin(x\pi/180)}{x\pi/180}$.

Finally, to get back to a form we know, let $y= x\pi/180$. Of course, as x goes to 0, so does y so we now have
$$\frac{\pi}{180}\lim_{y\to 0}\frac{sin(y)}{y}$$

lurflurf
Homework Helper
We can work without regard to angle measure system to find
$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x}=\frac{\pi}{\arccos(-1)}$$
Wolfram alpha humorously gives
$$\lim_{x \rightarrow 0} \frac{\sin(x^{\circ})}{x}=\, ^{\circ}$$
and
$$\lim_{x \rightarrow 0} \frac{\sin(x^{\circ})}{x ^{\circ}}=\frac{ ^{\circ}}{^{\circ}}=1$$
because it defined the constant
$$^{\circ}=\frac{\pi}{180}$$
and for a constant c (not zero in the second limit)
$$\lim_{x \rightarrow 0} \frac{\sin(c \, x)}{x}=\, c$$
and
$$\lim_{x \rightarrow 0} \frac{\sin(c \, x)}{c \, x }=1$$

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