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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 1

Lesson 15: Limits of trigonometric functions

# Trig limit using Pythagorean identity

In this video, we explore finding the limit as θ approaches 0 for the expression (1-cosθ)/(2sin²θ). By using the Pythagorean identity, we rewrite the expression to simplify it and avoid the indeterminate form 0/0. This allows us to evaluate the limit and find the answer, 1/4.

## Want to join the conversation?

• I'm still having confusion understanding why Sal changes the function from f(x) to g(x) at . Why does he do that? • besides using a cheat sheet and doing a ton of problems, is there a simple way to either remember or derive the double (and half- for that matter) angle identities. you know, something like the magic hexagon or a nice mnemonic? •   Found this when I was in 10th grade been using it since:

For sin(a+b)= sin(a)cos(b)+cos(a)sin(b)
Say out loud: "sine cosine cosine sine"

For cos(a+b)= cos(a)cos(b)-sin(a)sin(b)
Say out loud: "cosine cosine SIGN sine sine"
Notice the word "SIGN". The word SIGN is there to remind you that the SIGN of the rightmost term has the opposite SIGN of the input to the cosine on the left hand side of the equation (if its a+b, then subtract. if its a-b, then add).

Just say it out loud like 3 times. It has a really nice rythm:

"Sine Cosine Cosine Sine"
"Cosine Cosine SIGN Sine Sine"
"Sine Cosine Cosine Sine"
"Cosine Cosine SIGN Sine Sine"
"Sine Cosine Cosine Sine"
"Cosine Cosine SIGN Sine Sine"
• At Sal says that when we arrive at 0/0 we can still find the limit for this function, but if there was a different number in the numerator that is divided by zero that we would know that the limit for this function does not exist. Why is it that we know the limit does not exist when a non-zero number is divided by zero, but not when 0/0? Is this just because we can use the methods of rationalization and factorization on some expressions that work out to 0/0 but not on expressions that work out to non-zero/0? Or is there a deeper reason for this? • You will notice that many limit problems are in the form 0/0 and the techniques we use to evaluate these are by rationalizing or factoring and canceling. Later on we will learn about L'Hopital's rule as another technique. So we know that when it is 0/0 we know that it may or may not exist and we need to use one of the techniques to find out.

When the numerator is a non-zero, called it a, then a/0 is undefined (dividing by 0 is undefined), therefore its limit doesn't exist. In term of limit, a/0 = ±∞, depends on the sign. When a limit is ±∞, it is called infinite limit or unbound limit. In the infinite limit, the limit doesn't exist because we do not treat ±∞ as a number.
• At , I'm still a little confused... why can theta not equal zero if you cancel out the ( 1 - cos(theta))? • You want the functions to be equivalent to each other after you've simplified.

If you simplify the function by canceling out something in the denominator you'll need to watch out for values that would've made the denominator zero before you canceled them out, otherwise you would "create" a defined function value that wasn't there before.

Likewise if you try to rewrite an arbitrary function in a more complicated way and you create cases, where it is not defined, you'll need to add them somehow to make sure that the function stays exactly the same, here's an example:

if you want to rewrite the function f(x) = 1 in a more complicated way you could do something like f(x) = 1 = (x)^0 = (x)^(1-1) = x^(1) * x^(-1) = x/x but suddenly your function is not defined at x = 0 anymore because you'll get 0/0 (which is undefined since you can't divide by 0) instead of 1 as a result and you would have to add a condition that makes sure that your function is still defined at that point by writing f(x) = x/x for x not equal to 0, and f(x) = 1 for x = 0. (as piecewise defined function for example).

Basically you do this to stay "mathematically rigorous".
• Why does 0/0 mean indeterminate and not undefined? • When we cancel out the 1-cosx, shouldn't we also say that x cannot equal 180 as cos 180 will be -1 and 2(1+-1)(1-cosx) will make the denominator equal 0 • What should I watch to be brought up to speed on trig? I don’t remember anything other than sah cah toa • At , could someone please explain why 0/0 would not signify that there is no limit while any other number over 0 would signify that there is no limit? • Finding 0/0 means answering the question “what number times 0 is 0?”. Any number times 0 is 0, so any number can equal 0/0. So the value of 0/0 is inconclusive (or indeterminate), meaning that more work needs to be done to find the limit or determine that the limit doesn’t exist.

Let b represent a nonzero number. Finding b/0 means answering the question “what number times 0 is b”. No number times 0 is the nonzero number b, so no number can equal b/0. This means the limit definitely doesn’t exist.
• I understand 0/0 is an indeterminate form meaning we can find the actual limit using different methods, but how would you know if the limit is actually 0? • If at some point in the process of finding the limit, we find that the numerator goes to 0 while the denominator goes to a nonzero number, then the limit is 0.

This is not the only way the limit can be 0. For example, if, at some point in the process of finding the limit, we find that the numerator remains bounded while the denominator goes towards infinity or -infinity, then the limit is 0.
• at around , why does Sal say theta is not equal to zero? and then why does he change the function to g(x)? can someone please explain this part of the video? • There is a removable discontinuity at zero but you can find the limit by ignoring
the factor (1 - cos(theta)) in both the numerator and denominator. He's doing exactly the same you did when you had rational functions with removable discontinuities.
For example if you need the limit as x --> 1 of the function [ (x - 1) (x + 2) ] / [ (x - 1) (x + 3) ] you only need to find the limit as x --> 1 of the function (x + 2) / (x + 3) , which is doable by direct evaluation.
When you want to communicate this properly you give a different name for the two functions, say f for the first and g for the second. The difference between the two: you get in trouble by directly evaluating f at x=1 but you don't with g.
Now in trig the independent variable is frequently theta or some other cap from the greek well - but don't let that throw you. theta is just like x.
In Sal's example f cannot be evaluated directly at theta=0, but he made a new function by
"cancelling" the (1 - cos(theta)) factor from the numerator and denominator.
Hope this helps.