$$\lim_{x \to 0} \dfrac{\sin(1-\cos x)}{ x} $$
What is wrong with this argument: as $x$ approaches zero, both $x$ and $(1-\cos x)$ approaches $0$. So the limit is $1$ .
This is not about solving the limit because I already solved it but about the rate of both functions going to zero .
m referring to $$\lim_{x \to 0} \dfrac{\sin x}{ x} =1 $$
On
By your argument, $$\lim_{x\to 0}\frac{x}{x}$$ is also $0$ because as $x$ approaches zero, both $x$ and $x$ approach $0$, so the limit is $0$.
On
The argument is flawed in the sense that if both the numerator and denominator tends to 0 nothing much can be told about the limit. Remember $\frac{0}{0}$ is indeterminate. What matters is how fast each of them goes to 0.
In the question $1-\cos x= 2 \sin^2\frac{x}{2}$ goes to 0 much faster since for very small values of x $\sin x\sim x$, or $2 \sin^2\frac{x}{2} \sim 2(\frac{x}{2})^2=\frac{x^2}{2}$ Hence $\sin (1-\cos x)\sim \sin(\frac{x^2}{2}) \sim \frac{x^2}{2}$
Obviously $\frac{x^2}{2}$ goes to 0 much faster than $x$
On
The question is not whether the numerator goes to $0$ (which is what you show), but whether it goes to $0$ faster than the denominator.
To see this (without L'Hopital) consider the following:
By the Taylor series, $1-\cos(x)=O(x^2)$ as $x\to 0$.
Also, $\sin(x)=O(x)$, so $\sin(1-\cos(x))=O(x^2)$. This becomes a $O(x)$ when divided by the denominator $x$, so that the limit is $0$.
On
What is wrong with this argument ; as $x$ approaches zero , both $x$ and $(1-\cos x)$ approaches $0$ .
So the limit is 1.
What is wrong with this argument is that it is FALSE.
For example in $\lim_{x\to 0^+}\frac x{x^2}$ both $x$ and $x^2$ tend to zero, but the limit is $+\infty$.
As another example in $\lim_{x\to 0}\frac {x^2}x$ both $x^2$ and $x$ tend to zero, but the limit is $0$.
And for $\lim_{x\to 0}\frac {2x}x$ both $2x$ and $x$ tend to zero, but the limit is $2$.
So both numerator and denominator tending to zero DO NOT imply the fraction tends to $1$ or to any other pre-selected value.
On
You are correct that $$\lim_{x \to 0} \dfrac{\sin x}{ x} =1 .$$
This also implies that $$\lim_{x \to 0} \dfrac{\sin(1-\cos x)}{1-\cos x} = \lim_{(1-\cos x) \to 0} \dfrac{\sin(1-\cos x)}{1-\cos x} = 1 .$$
Therefore
\begin{align} \lim_{x \to 0} \frac{\sin(1-\cos x)}{x} &= \lim_{x \to 0} \left( \frac{\sin(1-\cos x)}{1-\cos x} \cdot \frac{1-\cos x}{x}\right)\\ &= \left( \lim_{x \to 0}\frac{\sin(1-\cos x)}{1-\cos x} \right) \cdot \left( \lim_{x \to 0} \frac{1-\cos x}{x} \right)\\ &= 1 \cdot \left( \lim_{x \to 0} \frac{1-\cos x}{x} \right), \end{align} provided that all those limits exist (which they do).
So the question now is, what is $$\lim_{x \to 0} \frac{1-\cos x}{x}?$$
Hint: it is not $1$.
The wrong thing is that both the numerator and denominator of your expression go to $0$ at the same time and then your limit assumes the form $\frac{0}{0}$ which is undefined and inconclusive.
RATE OF BOTH FUNCTIONS GOING TO ZERO
You basically need to calculate the derivative of the $2$ functions to see which one goes to $0$ earlier.
Now $$\frac{d}{dx}\left(\sin x\right)=\cos x$$ and $$\frac{d}{dx}\left( x \right)= 1$$
And we know that $\cos x \le 1$ So you know who goes to $0$ first.
ALTERNATE SOLUTION
However, better is to use proper formula for limits and solve it in this way: $$\lim_{x \to 0} \dfrac{\sin(1-\cos x)}{ x} $$ $$=\left[\lim_{(1-\cos x) \to 0} \frac{\sin(1-\cos x)}{(1-\cos x)}\right] \cdot \lim_{x \to 0} \frac{(1-\cos x)}{ x} $$ $$= 1\cdot \lim_{x \to 0} \frac{(1-\cos x)}{ x}$$
And the limit has a simpler shape and has the form $\frac{0}{0}$.
So better to apply L'Hospital's Rule.
$$ \lim_{x \to 0} \frac{(1-\cos x)}{ x}$$ $$= \lim_{x \to 0} \frac{\sin x)}{ 1 }$$ $$= 0$$
Hence you can say that the limit is $0$ by mathematical rigour.