Please help me in Evaluating limit of a function sinx^0/x

  • #1

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?
 

Answers and Replies

  • #2
35,439
7,308

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?
 
  • #3
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
 
  • #4
92
0
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?
 
  • #5
STEMucator
Homework Helper
2,075
140
Remember to check limits from BOTH sides in this case since really what you want is :

##sin(1) lim_{x→0} \frac{1}{x}##
 
  • #6
35,439
7,308
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?
 
  • #7
3,816
92
I think that OP meant x degrees (x°) but clarification is required. If it is x degrees, it is easy to evaluate.
 
  • #8
35,727
12,319
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.
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,847
966
The limit [itex]\lim_{x\to 0}\frac{sin(x)}{x}= 1[/itex] is only valid for x in radians so if kashan123999's "[itex]x^o[/itex]" really does mean that x is measured in degees, then we need to convert to radians. [itex]\pi[/itex] radians correspond to 180 degrees so "x" degrees is the same as [itex]x\pi/180[/itex] radians. That is [itex]sin(x^o)[/itex] would be equivalent to [itex]sin(x\pi/180)[/itex]. Now, the "x" outside the trig function is a number, not a measurement so is not in "radians" or "degrees". We have, so far, [itex]\lim_{x\to 0}\frac{sin(x\pi/180)}{x}[/itex].

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


Finally, to get back to a form we know, let [itex]y= x\pi/180[/itex]. Of course, as x goes to 0, so does y so we now have
[tex]\frac{\pi}{180}\lim_{y\to 0}\frac{sin(y)}{y}[/tex]
 
  • #10
lurflurf
Homework Helper
2,440
138
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
$$\lim_{x \rightarrow 0} \frac{\sin(c \, x)}{c \, x }=1$$
 
Last edited:

Related Threads on Please help me in Evaluating limit of a function sinx^0/x

  • Last Post
Replies
7
Views
9K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
17
Views
3K
Replies
20
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
7
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
21
Views
3K
Replies
2
Views
2K
Top