How to Apply Theorems to Evaluate Trigonometric Limits?

[Numnum]

Homework Statement

Use theorems to find the limit:

<br /> \lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))<br />

Homework Equations

Theorems like
f(x)=c is continuous
f(x)=x is continuous
\lim_{x\rightarrow 0} \cos(x)=1
\lim_{x\rightarrow 0} \sin(x)=0
\lim_{x\rightarrow a} \sin(x)=sin(a)
\lim_{x\rightarrow 0} \sin(x-a)=0

The Attempt at a Solution

I'm not sure where to start, but I looked at the last theorem and thought that since the limit of sin(x-a)=0, it would turn that whole part into 0, and therefore it would turn to arctan(0). Didn't seem correct, so I instead thought to simplify the sin(x-1)/x-1 part by letting x-1 equal another variable?

 

[Dick]

Homework Statement

Use theorems to find the limit:

<br /> \lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))<br />

Homework Equations

Theorems like
f(x)=c is continuous
f(x)=x is continuous
\lim_{x\rightarrow 0} \cos(x)=1
\lim_{x\rightarrow 0} \sin(x)=0
\lim_{x\rightarrow a} \sin(x)=sin(a)
\lim_{x\rightarrow 0} \sin(x-a)=0

The Attempt at a Solution

I'm not sure where to start, but I looked at the last theorem and thought that since the limit of sin(x-a)=0, it would turn that whole part into 0, and therefore it would turn to arctan(0). Didn't seem correct, so I instead thought to simplify the sin(x-1)/x-1 part by letting x-1 equal another variable?

Good idea! Let u=x-1. You should also have a theorem about the limit of sin(u)/u as u->0.

 

[STEMucator]

What is $lim_{x → 0} \frac{sin(x)}{x}$?

How does it relate to $lim_{x → 1} \frac{sin(x-1)}{x-1}$?

 

[Numnum]

Is this right?

So I have:


1. <br /> \lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))<br />


2. <br /> \lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(u)}{u}}))<br />


3. <br /> \lim_{x\rightarrow 1} \cos(arctan(1))<br />

because of the theorem: \lim_{x\rightarrow 0}({\frac{\sin(x)}{x}}))=1

4. \lim_{x\rightarrow 1} cos({\frac{π}{4}})

5. ={\frac{1}{√2}}

 

[Dick]

Is this right?

So I have:


1. <br /> \lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))<br />


2. <br /> \lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(u)}{u}}))<br />


3. <br /> \lim_{x\rightarrow 1} \cos(arctan(1))<br />

because of the theorem: \lim_{x\rightarrow 0}({\frac{\sin(x)}{x}}))=1

4. \lim_{x\rightarrow 1} cos({\frac{π}{4}})

5. ={\frac{1}{√2}}

Yes, and you are using your 'continuous function' theorems after you've worked out the sin(u)/u part, yes?

 

[HallsofIvy]

Personally, I would do it the other way around. Since cosine is continuous, for all x, \lim_{x\to 1} cos(f(x))= cos(\lim_{x\to 1} f(x)).

That is, from \lim_{x\to 1} cos(actan(\frac{sin(x- 1)}{x}) we look at \lim_{x\to 1}arctan(\frac{sin(x-1)}{x}). And since arctan is continuous for all x we look at \lim_{x\to 1}\frac{sin(x-1)}{x-1}. As you say, that last limit is 1 so we have cos(arctan(1))= cos(\pi/4)= \frac{\sqrt{2}}{2}