Solving limit with algebraic tools

[Yegor]

Hello!
<br /> \lim_{x\rightarrow 0} \frac{-x(1 -\cos x)}{\sin x - x}<br />
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.

 

[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.

 

[VietDao29]

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

?

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

 

[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??

 

[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 \sin x expansion. Here is my result:
<br /> \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}<br />
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?

 

[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 &gt; x - \frac{x ^ 3}{3!}
Or slightly differently:
\forall x \in ]0, \frac{\pi}{2} [ \ : \sin x - x + \frac{x ^ 3}{3!} &gt; 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&#039;&#039;&#039;(x) &gt; 0 that means f''(x) is increasing on that interval, f''(0) = 0 => \forall x \in ]0, \frac{\pi}{2} [ \ : f&#039;&#039;(x) &gt; 0, that again means f'(x) is increasing on that interval, f'(0) = 0 => \forall x \in ]0, \frac{\pi}{2} [ \ : f&#039;(x) &gt; 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) &gt; 0 (Q.E.D)
-------------------
You can do exactly the same to prove:
\forall x \in ]-\frac{\pi}{2}, 0 [ \ : \sin x &lt; x - \frac{x ^ 3}{3!}
\forall x \in ]0, \frac{\pi}{2} [ \ : \sin x &lt; x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!}
\forall x \in ]-\frac{\pi}{2}, 0 [ \ : \sin x &gt; 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.

 

[Yegor]

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

 

[NateTG]

Hello!
<br /> \lim_{x\rightarrow 0} \frac{-x(1 -\cos x)}{\sin x - x}<br />
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.

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