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What is $ \lim_{x \rightarrow 0} \frac{1 - \cos x}{x}$? A simple way to evaluate this limit is to substitute $0$ for $x$ in the numerator to obtain

$ \displaystyle \lim_{x \rightarrow 0} \frac{1 - 1}{x} = \lim_{x \rightarrow 0} ( \frac{1}{x} - \frac{1}{x} ) = \lim_{x \rightarrow 0} (0) = 0 $

since $ \frac{1}{x} - \frac{1}{x} = 0$ since one quantity subtracted from the same quantity is 0. This technique circumvents the problem of division by zero while utilizing the fact that $\cos(0)$ is known.

Christina Daniel
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    What do you mean by "Is there a reason why this technique is valid?" You are substituting $0$ for $x$ in only part of the expression and then continue to calculate with the remaining expression. That's not valid at all, even if the answer happens to be zero – imranfat Jan 01 '21 at 17:23
  • I am wondering why it is mathematically incorrect to substitute partially. – Christina Daniel Jan 01 '21 at 17:25
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    $$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x} \to \frac{1-1}{0} = \frac 00$$ which is indeterminate. You can't pick and choose only one x to replace with 0. Via L'Hopital's you have $$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x} \to \frac{1-1}{0} = \lim_{x\to 0}\frac{\sin x}{1} = 0$$ – amWhy Jan 01 '21 at 17:31
  • @amWhy the deleted answers do answer the question. The question is not 'what is this limit' but 'why can't I do it bit by bit.' $x/x$ makes it clear why not – Empy2 Jan 01 '21 at 17:35
  • No, they do not explain why it's wrong. Perhaps it could be $\lim_{x\to 0} \frac x0 \to \pm \infty$. – amWhy Jan 01 '21 at 17:36
  • @amWhy , we went in the direction the O.P went. Ofcourse, it also could be $\lim_{x \to 0}\dfrac{\cos x-1}{x}=\dfrac{\cos x-1}{0} \to \pm \infty$. – V.G Jan 01 '21 at 17:41

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A counterexample: $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac12,\quad\enspace\text{not }0.$$ Indeed $\;1-\cos x=2\sin^2\tfrac x2$, so $$\frac{1-\cos x} {x^2}= \frac{2\sin^2\frac x2}{4\bigl(\frac x2\bigr)^2}=\frac12\biggl(\underbrace{\frac{\sin\frac x2}{\frac x2}}_{\underset{\textstyle 1}{\downarrow}}\biggr)^2$$

Bernard
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@ChristinaDaniel OK, here is a counter example: Consider the expression $\frac{\sin 2x}{x}$ and let $x$ go to zero: The answer to this limit is $2$. Now consider the expression $\frac{\sin 2x-0}{x}$ for $x$ going to zero. The answer to this limit is still $2$. But $\sin0=0$ so we can now consider the expression $\frac{\sin 2x-x}{x}$, again with $x$ going to zero. But now this limit is $1$. So when you do a "partial" substitution, the answer changes. In other words, when you substitute for $x$, you need to do that for every $x$ in the expression.

V.G
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imranfat
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No, you cannot claim that $x=0$ in the numerator while $x\ne0$ at the denominator !

Using your method, a simple way to evaluate this limit is to substitute $0$ for $x$ in the denominator to obtain $$ \displaystyle \lim_{x \rightarrow 0} \frac{\cos x - 1}{0} =\lim_{x \rightarrow 0}\pm\infty$$ as the numerator is nonzero.

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Let $f(x) = \frac{1-\ln x}{e-x}$. We wish to find $\lim_{x\to e}f(x)$.

Using the proposed method would return the wrong answer.

enter image description here

David Diaz
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It's invalid.

You can not replace a variable with a constant in one part of an expression but leave it as a variable in another.

If you want to estimate a limit by replacing a variable with a constant you must replace it everywhere. If you do that you ge $\frac {1 - \cos 0}{0} = \frac 00$ and that doesn't help us at all.

We must assume $x \ne 0$ and if we replace it we must replace it with $x = h\ne 0$ and we get $\lim_{x\to 0} \frac {1-\cos x}x \approx \frac {1-\cos h}{h}$ and we can't replace $h$ with $0$ in the top and not the bottom because $h$ ISN"T $0$. And whatever the $x$ in the numerator is, the $x$ in the denominator must be the same thing.

.....

The reasoning of the error is that a little fudging in the top $x\approx 0$ means $\cos x \approx \cos 0$ won't affect much. But that is wrong. The fudging in the bottom makes a huge difference. $\frac 1x \not \approx \frac 10$. That's a no-no.

Complete no-no.

And completely invalid.

fleablood
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