Homework Help: Finding limits using theorems

1. May 27, 2013

Numnum

1. The problem statement, all variables and given/known data

Use theorems to find the limit:

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

2. Relevant equations

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

3. 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?

2. May 27, 2013

Dick

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

Last edited: May 27, 2013
3. May 27, 2013

Zondrina

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

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

4. May 27, 2013

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. $$={\frac{1}{√2}}$$

5. May 27, 2013

Dick

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

6. May 27, 2013

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