Question

simplify:
1/x+h - 1/x

Answer 1

\(\large \lim_{x \rightarrow 0} \left( \frac{1}{x+h} - \frac{1}{x} \right) \) ?

Answer 2

Sorry I'm having trouble typing up commands at the moment.

Answer 3

and for the limit, I mean \(\large \lim_{h \rightarrow 0}\)

Answer 4

Okay. if you mean lim h->0 of (1/(x+h) - 1/x)/h

Let's look at the numerator first:

1/(x+h) - 1/x

the only way to do this subtraction is to express these fractions with a common denominator.

At this level of math many people know:

1/a - 1/b = (b - a)/ab, and so will write this out directly, but by observing in classes (actually Utubes of class sessions) this is also a mystery to many students, so let's go about this the long way, and I apologize if you are too advanced for this.

(x+h) and x in general do not have a common factor, so we have to multiply the first fraction by x/x, and the second fraction by (x+h)/(x+h). In both cases this is the same as multiplying by 1, so it won't change anything:

(1/(x+h))(x/x) - (1/x)((x+h)/(x+h))

x/((x+h)(x)) - (x+h)/((x+h)(x))

now these fractions have a common denominator so we can do the subtraction:

x/((x+h)(x)) - (x+h)/((x+h)(x)) = (x - (x+h))/((x+h)(x)) = -h/((x+h)(x))

Now this was just the numerator of the fraction we were taking the limit of, so altogether we have:

lim h->0 of (1/(x+h) - 1/x)/x = lim h->0 [-h/((x+h)(x))]/h

Lo and behold, the h's cancel

lim h->0 [-1/((x+h)(x))]/1 = lim h->0 -1/((x+h)(x))

now we can substitute h=0 without fear of division by zero

lim h->0 -1/((x+h)(x)) = -1/x^2

This is of course the derivative of 1/x.

Once you learn the power rule, you will express 1/x as x^(-1), and you will know
that the derivative of x^n = nx^(n-1) so the derivative of x^(-1) is -x^(-2) which happens to be the same as -1/x^2 which is the answer we got the long way.