Alan

As follows:

.

Let all sides have length 1

Distance BD = sqrt(2)

Half of BD = sqrt(2)/2

cos(ABD) = (sqrt(2)/2)/1 

ABD = cos^-1(sqrt(2)/2) → 45 degrees

Rewrite your recursive equation as U_n = U_n+1 + 2

U₆ = 3

U₅ = 3 + 2 → 5

U₄ = 5 + 2 → 7

... etc.

No problem! Don't worry about it.

Like so:

\((\sqrt{n^3+7n^2}-\sqrt{n^3})/\sqrt n \rightarrow \sqrt{n^2+7n}-\sqrt{n^2}\\ \rightarrow (n^2+7n)^{1/2}-n \rightarrow n(1+7/n)^{1/2}-n \\ \rightarrow n(1+(1/2)(7/n)+(1/2)(-1/2)(1/2!)(7/n)^2+...)-n\\ \rightarrow n + 7/2 - 49/(8n) + ... - n \rightarrow 7/2 - 49/(8n) + ...\)

As all the higher order terms involve a power of n in the denominator, they go to zero as n goes to infinity, so we are left with just 7/2 as the limit.

Here's a graph:

Ok. x = 3 is a vertical line.  This has infinite slope, so is basically undefined.

(I repeat: a horizontal line has zero slope).

General equation of a straight line:  y = mx + c

1. 0 = -1*(-1) + c. Rearrange to find c

2. -2 = 3*6 + c. Again, rearrange to find c.

The slope of a horizontal line is well defined- it is zero.

I think what is wanted is as follows:

General equation of a straight line is y =mx + c

Here we know that this line goes through point (3, -2) so we can say

-2 = m*3 + c

This means c = -2 - 3m.  Put this in the equation for the line:

y = mx - 2 - 3m

or

y = m(x - 3) - 2

The slope m is undefined.

If the discontinuity occurs at, say, x = p, then integrate from a to p-eps, and add to the integral from p+eps to b, where a < p, b > p and eps is very small. If possible, do the integration symbolically, then take the limits of the results as eps tends to zero.

Incidentally, your f(x) example is continuous at x = 1: plot it to see!