Question

How do you use the limit definition of the derivative to find the derivative of `f(x)=x/(x+1)`?

Answer 1

See below for two possible solutions.

Explanation:

Soultion 1

**Using: ** `f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h`

`f'(x) = lim_(hrarr0)([(x+h)/(x+h+1)]-[x/(x+1)])/h`

If we try substitution, we get the indeterminate form `0/0`. Let's get a single fraction on top over `h/1`, then multiply the top by the reciprocal of the bottom (invert and multiply).

`= lim_(hrarr0)(((x+h)(x+1)-x(x+h+1))/((x+h+1)(x+1))) / (h/1)`

`= lim_(hrarr0)(((x^2+x+xh+h-x^2-xh-x))/((x+h+1)(x+1))) / (h/1)`

`= lim_(hrarr0)((h)/((x+h+1)(x+1))) / (h/1)`

`= lim_(hrarr0)((h)/(x+h+1)(x+1)) * 1/h`

`= lim_(hrarr0) 1/((x+h+1)(x+1))`.

Now, when we evaluate, we do not get `0` in the denominator, so the limit is

`= 1/(x+1)^2`.

`f'(x) = 1/(x+1)^2`.

Solution 2

Using `f'(x) = lim_(trarrx)(f(t)-f(x))/(t-x)`

`lim_(trarrx)(f(t)-f(x))/(t-x) = lim_(trarrx)([t/(t+1)]-[x/(x+1)])/(t-x)`

If we try substitution, we get the indeterminate form `0/0`. Let's get a single fraction on top over `(t-x)/1`, then multiply the top by the reciprocal of the bottom (invert and multiply).

`= lim_(trarrx)((t(x+1)-x(t+1))/((t+1)(x+1))) / ((t-x)/1)`

`= lim_(trarrx)(((tx+t-tx-x))/((t+1)(x+1))) / ((t-x)/1)`

`= lim_(trarrx)((t-x)/((t+1)(x+1))) / ((t-x)/1)`

`= lim_(trarrx)(t-x)/((t+1)(x+1)) * 1/(t-x)`

`= lim_(trarrx) 1/((t+1)(x+1))`.

Now, when we evaluate, we do not get `0` in the denominator, so the limit is

`= 1/(x+1)^2`.

`f'(x) = 1/(x+1)^2`.