Why is $ \lim_{x \to 0} \frac{x^2}{1-\cos(x)} $ not equal to 0?

Question

I was watching Calculus 1 lectures on Youtube, particularly Professor Leonard's, and I thought I would be able to answer a question on the board very simply, but I've reached quite an odd answer and I can't identify the faulty logic in my way of solving. Here is the question and my steps to solve it:

Evaluate: $$ \lim_{x \to 0} \frac{x^2}{1-\cos(x)} $$

My approach to solving this was to take $ x $ out as a factor, then deal with the two expressions as two limits, and eventually treat the latter as a reciprocal of the fundamental $\lim_{x \to 0} \frac{1-\cos x}{x}= 0$

Steps: $$ \lim_{x \to 0} \left[x \cdot \frac{x}{1-\cos(x)}\right] $$ $$ \lim_{x \to 0} x \cdot \lim_{x \to 0} \frac{x}{1-\cos(x)} $$ $$ \lim_{x \to 0} x \cdot \lim_{x \to 0} \left(\frac{1-\cos(x)}{x}\right)^{-1} $$ $$ \lim_{x \to 0} x \cdot \left[\lim_{x \to 0} \frac{1-\cos(x)}{x}\right]^{-1} $$ $$ 0 \cdot 0^{-1} = 0 $$

Now I am aware that my answer is mathematically wrong and that the correct limit is 2, but I can't deduce why by using my current knowledge of calculus. I would appreciate it if someone pointed out the reason I can not evaluate this limit this way.

Answer 1

$0/0$ is indeterminate. Use L'Hospital's rule: $$\lim_{x\to 0} \frac{x^2}{1-\cos x} = \lim_{x\to 0} \frac{2x}{\sin x} = \lim_{x\to 0} \frac{2}{\cos x} = \frac{2}{\cos 0} = 2$$

Answer 2

Generally speaking the "intuitive" rules for calculating limits such as

$$ \lim f(x)g(x)=\lim f(x) \cdot\lim g(x) $$ do only work if all limits exist. This is not the case here since the limit $$ \lim_{x\to0} \frac{x}{1-\cos(x)} $$ does not exist. That being said if you extend the real numbers with plus and minus infinity you can also extend the rules to some more cases e.g. $\infty \cdot a=\infty$ for $a\neq0$. However that is not the case for your specific example which reads $0\cdot\infty$ which is often referred to as "indeterminate form".

Please refer to the chapter "properties" in the following link:

https://en.m.wikipedia.org/wiki/Limit_of_a_function

Answer 3

Note that $1-\cos(x)=2\sin^2(x/2)$, so $$ \begin{align} \frac{x^2}{1-\cos(x)} &=\frac{x^2}{2\sin^2(x/2)}\\ &=\frac{4\,(x/2)^2}{2\sin^2(x/2)}\\ &=2\left(\frac{x/2}{\sin(x/2)}\right)^2 \end{align} $$ Now we can apply $\lim\limits_{x\to0}\frac{\sin(x)}x=1$.

Answer 4

Another method is the "series" method. As $x \to 0$, we have $$ \cos x = 1 - \frac{x^2}{2} + o(x^2) \\ 1 - \cos x = \frac{x^2}{2} + o(x^2) \\ \frac{x^2}{1 - \cos x} = \frac{x^2}{\frac{x^2}{2} + o(x^2)} = \frac{1}{\frac{1}{2} + o(1)} \rightarrow 2 $$

Answer 5

"My approach to solving this was to take x out as a factor, then deal with the two expressions as two limits"

You can not EVER do that. EVER. It's a rookie mistake but the values of $x$ and the values of $1-\cos x$ are not independent.

If you look at the behavior of $x^2$ and $1-\cos x$ where $x$ is near zero you can't consider the behavior and nearness of one without also considering the behavior and nearness of the other. So you can not considered thir limits separately.

Here is an example of way it doesn't work.

Take $\lim_{x\to 0} \frac {x}{2x}$. If you consider the limits separately you can try "doing the denominator first". SO as $x \to 0$ we have $\frac {x}{2x}\to \frac 0{2x} = 0$. Then after we do that we consider $2x\to 0$ we get $\frac 0{2x} = 0$ cosistantly so $0\to 0$ as $2x\to 0$. So $\lim_{x\to 0} \frac x{2x}= \lim_{\color{green}x\to 0}\lim_{\color{red}x \to 0}\frac {\color{red}x}{\color{green}{2x}}=\lim_{\color{green}x \to 0}\frac 0{2x}=\lim_{\color{green}x \to 0} 0 = 0$.

Or we not $2x \to 0$ and we do that "first" $\frac {x}{2x} =\frac {x}0 = \infty$ and $\infty \to \infty$ as $x^2 \to 0$ so $\lim_{x\to 0} \frac x{2x}= \lim_{\color{red}x\to 0}\lim_{\color{green}x \to 0}\frac {\color{red}x}{\color{green}{2x}}=\lim_{\color{red}x \to 0}\frac x{0}=\lim_{\color{green}x \to 0} \infty = \infty$.

But if we do them both "at the same time" we get $\lim_{x\to 0}\frac x{2x} = \lim_{x\to 0} \frac 12 = \frac 12$.

So which is it. Well whenever we reduce $\color{red} x$ to so something small such as $\color{red}x = \delta$ then we must also reduce $\color{green}{2x}$ to something equivalently small such as $\color{green}{2x} = 2\delta$. We can just make one of the small and leave the other one hanging around till we are ready for it.

So the first two are wrong.

.....

Okay.... you are asking "but what about the theorem that says if $\lim_{x\to a} f(x) = M$ and $\lim_{x\to a} g(x) = N$ then $\lim_{x\to a} f(x)g(x) = MN$. Isn't that actually saying we can do just said we can't do".

Welllll.... sort of. But not really. In this case $\lim_{x\to a}f(x) = M$ and $\lim_{x\to a} g(x) =N$ must actually be things where $MN$ is a value that makes sense. If $\lim_{x \to a}f(x) = 0$ and $\lim_{x\to a}g(x) =\infty$ then "$0\cdot \infty$" is meaningless garbage and it doesn't make any sense to talk of it.

This is what we can "indeterminate form".

(SUBTLE thing to note: When we say $\lim x f(x)g(x) = ML = \lim f(x)\lim g(x)$ we aren't "separating" the limits until after we get the result. We are not taking the limits one after another, we are still playing with the limits both and the same time. We just know ahead of time the result will hold. So we can focus on one of the other so long as we are aware the limits themselves are both occuring simultaneous.)

.....

Fortunately, for limits in indeterminate form we have l'hopitals rule.

If $\lim \frac {x}{1-\cos x}$ yiels $\frac {\lim x}{\lim (1-\cos x)} = \frac 00$ is in indeterminate form, we can sleep easy knowing

$\lim \frac {x}{1 - \cos x} = \lim\frac {derivative\ of\ x^2}{derivative\ of\ 1-\cos x} = \lim \frac {2x}{\sin x}= \frac {2}{1} = 2$.

Answer 6

This is NOT an answer in any rigorous form, but plotting the graph of the given function can offer some intuition as well-after you have dealt with the calculus (or if to your best efforts you didn't manage to deal with it).