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

  1. Aug 2, 2013 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. 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?
  2. jcsd
  3. Aug 2, 2013 #2


    Staff: Mentor

    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.
  4. Aug 2, 2013 #3
    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
  5. Aug 2, 2013 #4
    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?
  6. Aug 2, 2013 #5


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    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}##
  7. Aug 2, 2013 #6


    Staff: Mentor

    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.
  8. Aug 3, 2013 #7
    I think that OP meant x degrees (x°) but clarification is required. If it is x degrees, it is easy to evaluate.
  9. Aug 3, 2013 #8


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    2017 Award

    Staff: Mentor

    Not easier than the regular limit of sin(x)/x, I think. And it does not fit to the explanations in the first post.

    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.
  10. Aug 3, 2013 #9


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    Science Advisor

    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]
  11. Aug 3, 2013 #10


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    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
    $$\lim_{x \rightarrow 0} \frac{\sin(x^{\circ})}{x}=\, ^{\circ}$$
    $$\lim_{x \rightarrow 0} \frac{\sin(x^{\circ})}{x ^{\circ}}=\frac{ ^{\circ}}{^{\circ}}=1$$
    because it defined the constant
    and for a constant c (not zero in the second limit)
    $$\lim_{x \rightarrow 0} \frac{\sin(c \, x)}{x}=\, c$$
    $$\lim_{x \rightarrow 0} \frac{\sin(c \, x)}{c \, x }=1$$
    Last edited: Aug 3, 2013
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