Finding the Limit of a Trig Function with Another Trig Function Inside

[odmart01]

Homework Statement

lim t--> 0 cos(1-(sint/t)

Homework Equations

lim theta-->0 sin(theta)/theta =1

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.

 

[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]

Homework Statement

lim t--> 0 cos(1-(sint/t)

Homework Equations

lim theta-->0 sin(theta)/theta =1

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 can't 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 can't 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.