# Limit $\lim_{x \to 0} \dfrac{\sin(1-\cos x)}{ x}$

$$\lim_{x \to 0} \dfrac{\sin(1-\cos x)}{ x}$$

What is wrong with this argument: as $x$ approaches zero, both $x$ and $(1-\cos x)$ approaches $0$. So the limit is $1$ .

1. How can we prove that they approaches zero at same rate?
2. 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 $$\lim_{x \to 0} \dfrac{\sin x}{ x} =1$$

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What does this have to do with "infinite decimals"? – Spencer Jan 25 at 11:51
Changing the limit to 1 doesn't really help. The answer is indeed 0. – Shaswata Jan 25 at 11:53
Yes I know , but my question is both what we taking the sine of and x goes to 0 . so by that rule , $$\lim_{x \to 0} \dfrac{\sin x}{ x} =1$$ it should be 1. – Jacksoja Jan 25 at 11:56
I see. I hope your doubt has been cleared now – Shaswata Jan 25 at 11:58
Yes thank you all who took your time to help me :) – Jacksoja Jan 25 at 12:00

What is wrong with this argument ; as $x$ approaches zero , both $x$ and $(1-\cos x)$ approaches $0$ .

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.

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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 $0$.

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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 $x$

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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 $0$.

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You say "without l'Hopital", and then you go straight to a Taylor series argument. You do know that they are the same thing, right? – Arthur Jan 25 at 11:50
I'm fully aware, but the OP may not be.. @Arthur – Your Ad Here Jan 25 at 11:56

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.

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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 $1$.

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• in blue the plot of $\dfrac{\sin(x)}{x}$,
• in green that of $\dfrac{\sin(1-\cos(x))}{x}$.
In fact, for small $x$, $\sin(x)\approx x$ and $\cos(x)\approx1-\dfrac{x^2}2$, so that the first function is approximately $1$ and the second $\dfrac x2$.