Question

Is that any examples that can combine function, limit, differentiation and integration into one?

Answer 1

I'm not sure quite what you mean, but see below for my attempt at an answer (i.e. an example which incorporates limits, functions, derivatives, and integrals).

Explanation:

Here is my example:

Let `f(x) = secx`. Find `lim_(x rarr 0) int f(x) dx`

First comes the integration, in finding `int sec x dx`, which isn't obvious, but the standard 'trick' is to multiple by a fraction equal to one. Let `u=secx + tanx`. We want to multiply `f(x)` by `u/u` (i.e. one).

Hence we are finding `int secx * (secx + tanx)/(secx + tanx) dx` = `int (sec^2(x) + secxtanx)/(secx + tanx) dx`

It just so happens that if we differentiate `u`, we get the numerator of our integrand:

Using the chain rule `d/dx(secx) = d/dx((cosx)^-1)`.
This gives a derivative of `-(cosx)^-2 * -sinx` which is simply equal to `secxtanx`.

Using the product rule to differentiate `tanx`, we get:
`d/dx(tanx) = (cosx*cosx - (-sinx*sinx)) / cos^2x = 1/cos^2x = sec^2x`.

Hence `d/dx(u) = sec^2(x) + secxtanx`

Thus our integral becomes:

`int (du)/dx * 1/(secx + tanx) dx`
`= int 1/u du`
`= log_e(u)`
`= log_e|secx + tanx|`

(Remember `int 1/x dx = log_e|x|`).

Now to find `lim_(x rarr 0) int f(x) dx` which is simply equivalent to `lim_(x rarr 0) log_e(secx + tanx)`.

Doing a bit of rearrangement we get:

`log_e(lim_(x rarr 0) (1+sinx)/cosx)`

Easily enough we can simply let `x=0`
`:. sinx = 0` and `:. cosx = 1`

Hence it just becomes: `log_e1`

`=0`.

I hope this helps.