>>12840333You haven't graphed a line, you have calculated a distance, which is a quantity, between two lines using a limit, therefore
>Prove to me a limit is a line. still remains unanswered, as a line has not been graphed.
this is one way to graph it, its shit but i don't care
lets produce a hypotenuse for a right angle triage with the length and height both =1, with the x and y axis being the remaining two sides, we thus get f(x) = -x + 1
If we wanted to replicate this hypotenuse like op's pic, we will need to use multiple vertical and horizontal segments that have to be graphed independently
Each horizontal segment can be graphed as seen in pic rel 1
Where n is equal to halve the amount of segments there will be (it must be halve to accommodate both the horizontal and vertical segments). Also you can only have whole numbers, each line represents an equal segment of a whole. You can't have halve a segment, otherwise each segment isn't the same length, thus at some point, the distance between the hypotenuse line, and faux hypotenuse 'staircase' line, there will be a distance greater than 0, thus n is only in whole numbers.
Where N is equal to what segment number it is (only horizontal segments), and it ranges from each whole number from =0 to =n only in whole numbers for the same reasons. Thus as n increases, another N value is accommodated, thus another function exists to compose the 'staircase' exists, this is how the series of equations can make the faux hypotenuse.
pic rel for vert segments
Thus as n increases, N also increases and the faux line resembles the actual hypotenuse more accurately
addendum, as another N value is accommodated, a new equation would just exist, but the graphing software doesn't let me, hence the fN(x) f0(x), f1(x)..
If we make n approach -> infinity using a limit, the limit is still not being graphed. The limit is used to generate a quantity to substitute n. It is still n being graphed. A limit produces a value, its not a line.
