# Solving limit with algebraic tools

1. Nov 25, 2005

### Yegor

Hello!
$$\lim_{x\rightarrow 0} \frac{-x(1 -\cos x)}{\sin x - x}$$
I solved this limit using L'Hopital and expanding trigonometric functions to series.
But i have to solve it using algebraic tools (without series). I don't know how to do it. $$\sin x - x$$ looks difficult to deal with.

Last edited: Nov 25, 2005
2. Nov 25, 2005

### benorin

You have likely proved that

$$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$$ (EDIT: yes, I meant 0, not $\infty$, thanks)

If so, use it.

Last edited: Nov 26, 2005
3. Nov 25, 2005

### VietDao29

???

Are you sure the limit above is correct? Let's check it again.
$$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$$
Or
$$\lim_{x\rightarrow \infty} \frac{\sin(x)}{x}=0$$

4. Nov 26, 2005

### Yegor

Possibly, Benorin meant $$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$$,
but i believe, that this can't help in solving initial problem. I have to use further terms of sinx series. $$\sin x = x - \frac{x^3}{3!}+...$$ But how can i get them algebraically??

5. Nov 28, 2005

### Yegor

Realy, nobody have any idea about initial problem?

6. Nov 28, 2005

### matt grime

there aren't many things that spring to mind except using your trig identities.

7. Nov 28, 2005

### Yegor

So, i tried to get some more terms of $$\sin x$$ expansion. Here is my result:
$$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} = 2 \sin \frac{x}{2} (\cos \frac{x}{2} -1 +1) = -2 \sin \frac{x}{2} (1-\cos \frac{x}{2}) + 2\sin \frac{x}{2} = -4 \sin \frac{x}{2} (\sin \frac{x}{4})^2 +2\sin \frac{x}{2}$$
Now if x->0 i got that $$\sin x = x - \frac{x^3}{8}+...$$ and not $$\sin x = x - \frac{x^3}{3!}+...$$ as it should be. Where do i went wrong?

Last edited: Nov 28, 2005
8. Nov 29, 2005

### VietDao29

If you drop off all of the terms of power higher than 1 in 2 sin(x / 2), you must as well drop off all the term has the poer higher than 1 in the other terms.
$$-4\sin \frac{x}{2} \sin ^ 2 \frac{x}{4} = -\frac{x ^ 3}{8}$$ should be also dropped off, as the degree of x is 3 (which is obviously greater than 1).
That'll give:
$$\sin x = x + ...$$, and that's correct.
--------------------
This is my approach:
I'll prove
$$\forall x \in ]0, \frac{\pi}{2} [ \ : \sin x > x - \frac{x ^ 3}{3!}$$
Or slightly differently:
$$\forall x \in ]0, \frac{\pi}{2} [ \ : \sin x - x + \frac{x ^ 3}{3!} > 0$$
Let f(x) := sin x - x + x3 / 3!
f'(x) = cos x - 1 + x2 / 2!
f''(x) = -sin x + x
f'''(x) = -cos x + 1
$$\forall x \in ]0, \frac{\pi}{2} [ \ : f'''(x) > 0$$ that means f''(x) is increasing on that interval, f''(0) = 0 => $$\forall x \in ]0, \frac{\pi}{2} [ \ : f''(x) > 0$$, that again means f'(x) is increasing on that interval, f'(0) = 0 => $$\forall x \in ]0, \frac{\pi}{2} [ \ : f'(x) > 0$$.
That means f(x) is increasing on that interval, f(0) = 0, so we have:
$$\forall x \in ]0, \frac{\pi}{2} [ \ : f(x) > 0$$ (Q.E.D)
-------------------
You can do exactly the same to prove:
$$\forall x \in ]-\frac{\pi}{2}, 0 [ \ : \sin x < x - \frac{x ^ 3}{3!}$$
$$\forall x \in ]0, \frac{\pi}{2} [ \ : \sin x < x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!}$$
$$\forall x \in ]-\frac{\pi}{2}, 0 [ \ : \sin x > x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!}$$
-------------------
It means:

$$\forall x \in ]0, \frac{\pi}{2} [ \ : -\frac{x ^ 3}{3!} \leq \sin x - x \leq -\frac{x ^ 3}{3} + \frac{x ^ 5}{5!}$$
$$\forall x \in ]-\frac{\pi}{2}, 0 [ \ : -\frac{x ^ 3}{3!} \geq \sin x - x \geq -\frac{x ^ 3}{3} + \frac{x ^ 5}{5!}$$
From there, just use Squeeze theorem, to evaluate the limit of:
$$\lim_{x \rightarrow 0} \frac{\sin x - x}{x ^ 3}$$.
And, that's way longer than to use L'Hopital rules, or sin expansion. Maybe someone will come up with something shorter.

Last edited: Nov 29, 2005
9. Nov 29, 2005

### Yegor

Thank you very much, VietDao. Your approach is realy very nice. But unfortunately i cannot use derivatives . Just algebra:yuck:

10. Nov 29, 2005

### NateTG

Try
$$\frac{-x(1 -\cos x)}{\sin x - x} \times \frac{1+\cos x}{1+\cos x}$$
and some trig identities.