Solving limit with algebraic tools

[Yegor]

Hello!
[tex]
\lim_{x\rightarrow 0} \frac{-x(1 -\cos x)}{\sin x - x}
[/tex]
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. [tex]\sin x - x[/tex] looks difficult to deal with.

 

[benorin]

You have likely proved that

[tex]\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1[/tex] (EDIT: yes, I meant 0, not [itex]\infty[/itex], thanks)

If so, use it.

 

[VietDao29]

You have likely proved that
[tex]\lim_{x\rightarrow\infty} \frac{\sin(x)}{x}=1[/tex]
If so, use it.

?

Are you sure the limit above is correct? Let's check it again.
Shouldn't it read:
[tex]\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1[/tex]
Or
[tex]\lim_{x\rightarrow \infty} \frac{\sin(x)}{x}=0[/tex]

 

[Yegor]

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

 

[Yegor]

Realy, nobody have any idea about initial problem?

 

[matt grime]

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

 

[Yegor]

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

 

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

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

 

[Yegor]

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

 

[NateTG]

Hello!
[tex]
\lim_{x\rightarrow 0} \frac{-x(1 -\cos x)}{\sin x - x}
[/tex]
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. [tex]\sin x - x[/tex] looks difficult to deal with.

Try
[tex]\frac{-x(1 -\cos x)}{\sin x - x} \times \frac{1+\cos x}{1+\cos x}[/tex]
and some trig identities.