# Evaluating Trigonometric Limits

• kwikness
In summary: Just the fact that the op is asking this question, shows that he is not yet familiar with the things that you are talking about.In summary, the expression sin(0)/0 does not have a defined value, but when taking the limit as x approaches 0, it becomes 1. This can be demonstrated using various methods, such as using the unit circle or looking at the Maclaurin series for sin(x). However, for a simpler explanation, one can take the derivative of the numerator and denominator separately and then plug in 0 for x to reach the answer of 1.
kwikness
http://calcchat.tdlc.com/solutionart/calc8e/01/c/se01c01067.gif

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

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kwikness said:
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.

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Look the function sin(x)/x si undefined for x=0, so you have to take the limit as x-->0.

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

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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

Nevetsman said:
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.

Another way to show that

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

is to look at the maclaurin series for sinx

Nevetsman said:
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!

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

## 1. What are trigonometric limits?

Trigonometric limits are limits that involve trigonometric functions, such as sine, cosine, and tangent. They are used to determine the behavior of these functions as they approach a certain value.

## 2. How do you evaluate a trigonometric limit?

To evaluate a trigonometric limit, you can use algebraic manipulation, trigonometric identities, and knowledge of the properties of limits. You can also use a graphing calculator or a table of values to approximate the limit.

## 3. What are the common trigonometric limit formulas?

Some common trigonometric limit formulas include:
- lim sin(x)/x = 1
- lim (1-cos(x))/x = 0
- lim (sin(x))^(1/x) = 1
- lim (1+1/x)^x = e
- lim (1-cos(x))/x^2 = 1/2
- lim (1+tan(x))/x = 1
- lim (sin(x)/x)^x = 1

## 4. What are the challenges in evaluating trigonometric limits?

Some challenges in evaluating trigonometric limits include:
- Identifying the correct trigonometric identity or formula to use
- Dealing with indeterminate forms, such as 0/0 or infinity/infinity
- Understanding the behavior of trigonometric functions at certain values, such as 0, π/2, and π

## 5. What are some real-life applications of trigonometric limits?

Trigonometric limits are used in many fields, including engineering, physics, and astronomy. Some real-life applications include calculating the maximum height of a projectile, determining the stability of structures, and analyzing the motion of pendulums and waves.

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