I found a copy of the relevant passage from Berkeley's works at [this web site][1]. I have cut and pasted from that site, and I have reformatted the mathematics; apologies to the good Bishop for any alterations in meaning.


>XIV. To make this Point plainer, I shall unfold the reasoning, and propose it in a fuller light to your View. It amounts therefore to this, or may in other Words be thus expressed. I suppose that the Quantity $x$ flows, and by flowing is increased, and its Increment I call $o$, so that by flowing it becomes $x + o$. And as $x$ increaseth, it follows that every Power of $x$ is likewise increased in a due Proportion. Therefore as $x$ becomes $x + o$, $x^n$ will become $(x + o)^n$: that is, according to the Method of infinite Series,
$$x^n + nox^{n-1} + \frac{n^2-n}{2} o^2 x^{n-2} + \text{  etc.}$$
And if from the two augmented Quantities we subduct the Root and the Power respectively, we shall have remaining the two Increments, to wit,
$$o  \text{  and  }  nox^{n-1} + \frac{n^2-n}{2} o^2 x^{n-2} + \text{  etc.}$$
which Increments, being both divided by the common Divisor o, yield the Quotients
$$1 \text{  and  } nx^{n-1} + \frac{n^2-n}{2} ox^{n-2} + \text{  etc.}$$ 
which are therefore Exponents of the Ratio of the Increments. Hitherto I have supposed that $x$ flows, that $x$ hath a real Increment, that $o$ is something. And I have proceeded all along on that Supposition, without which I should not have been able to have made so much as one single Step. From that Supposition it is that I get at the Increment of $x^n$, that I am able to compare it with the Increment of $x$, and that I find the Proportion between the two Increments. I now beg leave to make a new Supposition contrary to the first, i. e. I will suppose that there is no Increment of $x$, or that $o$ is nothing; which second Supposition destroys my first, and is inconsistent with it, and therefore with every thing that supposeth it. I do nevertheless beg leave to retain $nx^{n - 1}$, which is an Expression obtained in virtue of my first Supposition, which necessarily presupposeth such Supposition, and which could not be obtained without it: All which seems a most inconsistent way of arguing, and such as would not be allowed of in Divinity. 

It looks to me that the Berkeley's argument amounts to an argument raised by every discerning student in a nonrigorous first semester calculus course: ``Is the increment zero? or not zero? How can it be both? That's inconsistent!'' In which case I would invite the good Bishop to come to my office hours where I would introduce him to $\epsilon$, $\delta$ proofs. 

I bet I could even convince the Bishop that Divinity would allow it: "Suppose the Devil gives you any $\epsilon > 0$. This $\epsilon$, although positive, might be very, very, very small, as small as the Devil likes...".


  [1]: http://www.maths.tcd.ie/pub/HistMath/People/Berkeley/Analyst/Analyst.html