Is there a reason why this technique is invalid?  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.
 A: 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
$$\frac{1-\cos x} {x^2}= \frac{2\sin^2\frac x2}{4\bigl(\frac x2\bigr)^2}=\frac12\biggl(\underbrace{\frac{\sin\frac x2}{\frac x2}}_{\underset{\textstyle 1}{\downarrow}}\biggr)^2$$
A: @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.
A: 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.
A: 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.

A: 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.
