differentiate the function f(x) = 2/x using the limit defintion?

I'll change your "trianglex" to an h.

The definition that I use is:

lim h->0 of [f(x + h) - f(x)] / h

You are given:

f(x) = 2 / x

so:

f(x + h) = 2 / (x + h)

If we substitute the above expressions into the limit equation we get:

lim h->0 of [f(x + h) - f(x)] / h

lim h->0 of [2 / (x + h) - 2 / x] / h

If we try to substitute 0 in for h we have 0 in the denominator, so we need to do something about that.   I'll start with simplifying the difference of two fractions in the numerator, starting with getting a common denominator:

lim h->0 of {2x / [x(x + h)] - 2(x + h) / [x(x + h)]} / h

Now we can subtract the numerators:

lim h->0 of {[2x - 2(x + h)] / [x(x + h)]} / h

And simplify both halves of the fraction that's in the numerator of the larger fraction:

lim h->0 of [(2x - 2x - 2h) / (x² + hx)] / h

the x terms cancel out in the numerator's numerator:

lim h->0 of [-2h / (x² + hx)] / h

Now we can turn the division of fractions into the multiplication of the reciprocal:

lim h->0 of [-2h / (x² + hx)] * 1 / h

lim h->0 of -2h / [h(x² + hx)]

We can now cancel the common factor "h" in the remaining fraction:

lim h->0 of -2 / (x² + hx)

Now we can safely substitute 0 in for h to get the derivative of your original function:

-2 / (x² + 0x)

-2 / (x² + 0)

-2 / x²

That's your answer.

First, sate the limit definition here.

Then we will show you how to apply it.