Q: Can't properly evaluate a limit algebraically. I tried my best to evaluate this as the  limit approaches zero. The textbook says it should approach two but I get zero.  $$\frac{2(x+\Delta x)-2x}{\Delta x} =\frac{2x+2\Delta x-2x}{\Delta x}=\frac{ 2\Delta x}{\Delta x}=\Delta x $$
Any help would be awesome.

A: We start with the expression 
$$\frac{2(x+\Delta x)-2x}{\Delta x}$$
$$=\frac{2x+2\Delta x-2x}{\Delta x}$$
$$=\frac{2\Delta x}{\Delta x}$$
$$=2\frac{\Delta x}{\Delta x}$$
At this point we have an infinitesimal over itself, and we know have the definition that $\frac{a}{a} = 1$
$$=2(1)$$
$$=2$$

A: You canceled the wrong thing out in the last step. You wrote $\frac{2 \Delta x}{\Delta x} = \Delta x$. Instead, the $\Delta x$'s cancel and you get 2.

A: @JoshuaMorris, you have it correct EXCEPT for the last step where you subtract the denominator from the numerator. You just need to divide both by the common factor i.e. DELTA x and you will get 2.
