How to Apply Theorems to Evaluate Trigonometric Limits?

[Numnum]

Homework Statement

Use theorems to find the limit:

$$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))$$

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:

$$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))$$

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. $$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))$$


2. $$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(u)}{u}}))$$


3. $$\lim_{x\rightarrow 1} \cos(arctan(1))$$

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. $$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))$$


2. $$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(u)}{u}}))$$


3. $$\lim_{x\rightarrow 1} \cos(arctan(1))$$

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 $cos(arctan(1))= cos(\pi/4)= \frac{\sqrt{2}}{2}$