Evaluating Trigonometric Limits

[kwikness]

Hi, I had a question about the following problem:
http://calcchat.tdlc.com/solutionart/calc8e/01/c/se01c01067.gif

Can someone explain how sin(0)/0 = 1?

 

[sutupidmath]

Hi, I had a question about the following problem:
http://calcchat.tdlc.com/solutionart/calc8e/01/c/se01c01067.gif

Can someone explain how sin(0)/0 = 1?

No one can explain to you what you are asking for because it is not true. It is true only when we take the limit as x goes to zero.

 

[sutupidmath]

Look the function sin(x)/x si undefined for x=0, so you have to take the limit as x-->0.
use the following link it will be very helpful

http://www.csun.edu/~ac53971/courses/math350/xtra_sine.pdf

 

[Nevetsman]

Hello!

I'm new to this forum, and was passing through the threads when I found your post. I have a hint!

In fact, the lim(x->0) of sin(x)/x is 1! When finding limits the first thing to do is plug in the limit to observe how the function behaves. In this situation, when you plug 0 in for 'x' you get 0/0. Incidentally, this is an indeterminate form, just like infinity/infinity.

When limits result in an indeterminate form the solution can be found by taking the derivative of the numerator and denominator of the function. DO NOT perform the quotient rule to find this derivative. Simply take the derivative of the numerator separately from the denominator, then plug in 0 for 'x' again. You should reach the answer you were looking for.

Steve

 

[sutupidmath]

Hello!

I'm new to this forum, and was passing through the threads when I found your post. I have a hint!

In fact, the lim(x->0) of sin(x)/x is 1! When finding limits the first thing to do is plug in the limit to observe how the function behaves. In this situation, when you plug 0 in for 'x' you get 0/0. Incidentally, this is an indeterminate form, just like infinity/infinity.

When limits result in an indeterminate form the solution can be found by taking the derivative of the numerator and denominator of the function. DO NOT perform the quotient rule to find this derivative. Simply take the derivative of the numerator separately from the denominator, then plug in 0 for 'x' again. You should reach the answer you were looking for.

Steve

'
well, no offends, but i think that the op is not only interested to know how to reach the final answer, rather he wants to know why lim(x-->0)sin(x)/x =1
and the best way to know why such a thing is true is to go back to its very deffinition. I mean to use the unit circle and construct three triangles in it. The link that i posted in post #3 explains the whole thing. I also think that the op haven't yet reached the point of using l'hopital rule, and to know why the l'hopital really works it needs quite some rigorous calculus, which the op might lack at the moment.

 

[rock.freak667]

Another way to show that

lim$$_{x\rightarrow 0} \frac{sinx}{x}=1$$

is to look at the maclaurin series for sinx

 

[sutupidmath]

Hello!

I'm new to this forum, and was passing through the threads when I found your post. I have a hint!

In fact, the lim(x->0) of sin(x)/x is 1! When finding limits the first thing to do is plug in the limit to observe how the function behaves. In this situation, when you plug 0 in for 'x' you get 0/0. Incidentally, this is an indeterminate form, just like infinity/infinity.

When limits result in an indeterminate form the solution can be found by taking the derivative of the numerator and denominator of the function. DO NOT perform the quotient rule to find this derivative. Simply take the derivative of the numerator separately from the denominator, then plug in 0 for 'x' again. You should reach the answer you were looking for.

Steve

Welcome to PF, by the way!

 

[sutupidmath]

For gods sake, the op is asking for a simple answer, but you are rather making an elefant out of a fly!