# Limit of trig function

#### odmart01

1. The problem statement, all variables and given/known data
lim t--> 0 cos(1-(sint/t)

2. Relevant equations
lim theta-->0 sin(theta)/theta =1

3. The attempt at a solution
I usually don't have a problem with these limits, but I've never done 1 with a trig function inside another trig function. So I don't know how to begin this one.

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#### Inferior89

If f(x) --> A when x --> a and g(x) --> B when x --> A then g(f(x)) --> B when x --> a, if a is in the domain of g(x).

Here your f(x) = sin(x)/x and g(x) = cos(1-x) so g(f(x)) = cos(1-sin(x)/x) and your a is 0.

#### Oerg

have you heard of l'Hospital's rule?

#### HallsofIvy

1. The problem statement, all variables and given/known data
lim t--> 0 cos(1-(sint/t)

2. Relevant equations
lim theta-->0 sin(theta)/theta =1

3. The attempt at a solution
I usually don't have a problem with these limits, but I've never done 1 with a trig function inside another trig function. So I don't know how to begin this one.
You are missing parentheses. Is this "cos(1)- (sin(t)/t)" or is it "cos(1- (sin(t)/t))"?

If it is the first, then cos(1) is just a constant: the limit is cos(1)- 1.

If it is the second, then use the fact that cosine is continuous: the limit is cos(1- 1)= cos(0)= 1.

#### odmart01

You are missing parentheses. Is this "cos(1)- (sin(t)/t)" or is it "cos(1- (sin(t)/t))"?

If it is the first, then cos(1) is just a constant: the limit is cos(1)- 1.

If it is the second, then use the fact that cosine is continuous: the limit is cos(1- 1)= cos(0)= 1.
it is cos(1- (sin(t)/t)), but what do you do with the (sin(t)/t) in the inside, you cant just assume its 0 because then it does not exist. How are you getting 1-1. Can you explain?

#### Inferior89

it is cos(1- (sin(t)/t)), but what do you do with the (sin(t)/t) in the inside, you cant just assume its 0 because then it does not exist. How are you getting 1-1. Can you explain?
It exists.

You said yourself in the "relevant equations" part that:
lim theta-->0 sin(theta)/theta =1
which is the same as
lim t -->0 sin(t)/t = 1

#### odmart01

It exists.

You said yourself in the "relevant equations" part that:
lim theta-->0 sin(theta)/theta =1
which is the same as
lim t -->0 sin(t)/t = 1
Ohh. I feel so stupid.

"Limit of trig function"

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