Problem: Show 1 = 2 (with calculus)
“Solution”: Consider the following:
$ 1^2 = 1$
$ 2^2 = 2 + 2$
$ 3^2 = 3 + 3 + 3$
$ x^2 = x + x + \dots + x$ ($ x$ times)
And since this is true for all values of $ x$, we may take the derivative of both sides, and the equality remains true. In other words:
$ 2x = 1 + 1 + \dots + 1$ ($ x$ times)
Which simplifies to $ x=2x$, and plugging in $ x=1$ we have $ 1 = 2$, as desired.
Explanation: Though there are some considerations about the continuity of adding something to itself a variable number of times, the true error is as follows. If we are taking the derivative of a function with respect to $ x$, then we need to take into account all parts of that function which involve the variable. In this case, we ignored that the number of times we add $ x$ to itself depends on $ x$. In other words, $ x + x + \dots + x$ ($ x$ times) is a function of two variables in disguise:
$ f(u,v) = u + u + \dots + u$ ($ v$ times)
And our mistake was to only take the derivative with respect to the first variable, and ignore the second variable. Unsurprisingly, we made miracles happen after that.
Addendum: Continuing with this logic, we could go on to say:
$ x = 1 + 1 + \dots + 1$ ($ x$ times)
But certainly the right hand side is not constant with respect to $ x$, even though each term is.