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== 3.2#h(0.6em)Limits Along Different Roads


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Objectives]


      - Test a two-variable limit along straight approach paths and detect path-dependence.
- Compute the constant height of the lifted road along y = kx and match it to the scene.
- State what continuity at a point would demand of every approach path.
]


    In one variable, a point can be approached from only two sides, and a limit exists when left and right agree. In the plane there are infinitely many roads into any point — every direction, every curve — and #math.equation(block: false, alt: "lim sub open parenthesis x , y close parenthesis → open parenthesis a , b close parenthesis f open parenthesis x , y close parenthesis")[$lim_(( x , y ) → ( a , b )) f ( x , y )$] exists only if #emph[all] of them agree. That is a far stronger demand, and this section builds a surface on which you can watch it fail: different straight roads into the origin genuinely arrive at different heights.


    The specimen is


    #math.equation(block: true, alt: "f open parenthesis x , y close parenthesis equals the fraction 2 x y over x squared plus y squared ,")[$f ( x , y ) = frac(2 x y, x^(2) + y^(2)) ,$]


    defined everywhere except the origin. Away from #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] its graph is smooth ridges and valleys, but at the origin it pleats into a sharp vertical fin — and the fin, it will turn out, #emph[is] the failed limit.


    
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      The red-and-blue surface #math.equation(block: false, alt: "z equals 2 x y / open parenthesis x squared plus y squared close parenthesis")[$z = 2 x y / ( x^(2) + y^(2) )$] (the expression carries a tiny #math.equation(block: false, alt: "plus 0.000001")[$+ 0.000001$] in the denominator so the mesh stays finite at the origin), pleating into a vertical fin at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]. Three thick straight roads ride the surface through the origin: green along #math.equation(block: false, alt: "y equals 0")[$y = 0$] at height #math.equation(block: false, alt: "0")[$0$], red along #math.equation(block: false, alt: "y equals x")[$y = x$] at height #math.equation(block: false, alt: "1")[$1$], and blue along #math.equation(block: false, alt: "y equals minus x")[$y = − x$] at height #math.equation(block: false, alt: "minus 1")[$− 1$]. A fourth, amber road along #math.equation(block: false, alt: "y equals 2 x")[$y = 2 x$] hides in the object list.
    
    
=== Explore


    + Orbit slowly around the origin. Each colored road lies in a vertical plane over a straight line in the #math.equation(block: false, alt: "x y")[$x y$]-plane. For each of the three visible roads, read off two things: the line it travels over, and the height it travels at.
+ Every road is #emph[perfectly level] — constant height — yet all three pass over the same point #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]. What are the three heights over the origin? Could any single value #math.equation(block: false, alt: "f open parenthesis 0 , 0 close parenthesis")[$f ( 0 , 0 )$] agree with all three roads at once?
+ The hidden amber road runs along #math.equation(block: false, alt: "y equals 2 x")[$y = 2 x$], a line that sits #emph[between] #math.equation(block: false, alt: "y equals 0")[$y = 0$] (green, height #math.equation(block: false, alt: "0")[$0$]) and #math.equation(block: false, alt: "y equals x")[$y = x$] (red, height #math.equation(block: false, alt: "1")[$1$]). Before revealing it, predict: will its height land between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "1")[$1$]? Closer to which? Commit to a number.
+ Reveal the amber road with its eye toggle in the object list, orbit to a side view, and read its height against the grid. How close was your guess? The example below does the algebra.
+ Now make the scene grade your computation. Edit the amber road's #math.equation(block: false, alt: "y")[$y$]-component from 2\*t to 5\*t: the road jumps off the surface, because its #math.equation(block: false, alt: "z")[$z$]-component still carries the old height. Compute the correct height for slope #math.equation(block: false, alt: "5")[$5$], retype the #math.equation(block: false, alt: "z")[$z$]-component, and watch — if the road settles back onto the surface, your value is right.
+ All your computations follow one formula: the road over #math.equation(block: false, alt: "y equals k x")[$y = k x$] rides at height #math.equation(block: false, alt: "2 k / open parenthesis 1 plus k squared close parenthesis")[$2 k / ( 1 + k^(2) )$]. Which slope rides highest? Which lowest? Sketch height against #math.equation(block: false, alt: "k")[$k$] on paper.


    
=== Why the limit fails


    Along any line through the origin the function is constant, so each road is level — but the constants #emph[differ from road to road], filling the whole range from #math.equation(block: false, alt: "minus 1")[$− 1$] to #math.equation(block: false, alt: "1")[$1$]. For the limit to exist, one number would have to be the destination of every road at once. No number is, so


    #math.equation(block: true, alt: "lim sub open parenthesis x , y close parenthesis → open parenthesis 0 , 0 close parenthesis the fraction 2 x y over x squared plus y squared does not exist.")[$lim_(( x , y ) → ( 0 , 0 )) frac(2 x y, x^(2) + y^(2)) #h(1em) "does not exist."$]


    Continuity at the origin is then beyond rescue: it would require #math.equation(block: false, alt: "f open parenthesis 0 , 0 close parenthesis")[$f ( 0 , 0 )$] to equal a limit that is not there. No choice of value plugs the fin — the fin is the disagreement between roads, made solid.


    #examplebox("Example 1")[Every straight road has its own altitude][
Find the height at which the surface carries the road along #math.equation(block: false, alt: "y equals k x")[$y = k x$], then verify the amber road's height and compute the correct height after its slope is edited to #math.equation(block: false, alt: "5")[$5$].


      Substitute #math.equation(block: false, alt: "y equals k x")[$y = k x$] with #math.equation(block: false, alt: "x not equal to 0")[$x ≠ 0$]:


        #math.equation(block: true, alt: "f open parenthesis x , k x close parenthesis equals the fraction 2 x open parenthesis k x close parenthesis over x squared plus k squared x squared equals the fraction 2 k over 1 plus k squared ,")[$f ( x , k x ) = frac(2 x ( k x ), x^(2) + k^(2) x^(2)) = frac(2 k, 1 + k^(2)) ,$]


        a constant — the same for every #math.equation(block: false, alt: "x not equal to 0")[$x ≠ 0$], which is why each lifted road is level. For #math.equation(block: false, alt: "k equals 2")[$k = 2$] the height is #math.equation(block: false, alt: "the fraction 4 over 5 equals 0.8")[$frac(4, 5) = 0.8$], matching the amber road. For #math.equation(block: false, alt: "k equals 5")[$k = 5$] it is #math.equation(block: false, alt: "the fraction 10 over 26 equals the fraction 5 over 13 approximately equals 0.385")[$frac(10, 26) = frac(5, 13) ≈ 0.385$] — the value that reseats the edited road. The extremes come from #math.equation(block: false, alt: "open parenthesis 1 ∓ k close parenthesis squared greater than or equal to 0")[$( 1 ∓ k )^(2) ≥ 0$], which rearranges to #math.equation(block: false, alt: "| 2 k | less than or equal to 1 plus k squared")[$| 2 k | ≤ 1 + k^(2)$]: the height #math.equation(block: false, alt: "2 k / open parenthesis 1 plus k squared close parenthesis")[$2 k / ( 1 + k^(2) )$] never leaves #math.equation(block: false, alt: "[ minus 1 , 1 ]")[$[ − 1 , 1 ]$], and touches #math.equation(block: false, alt: "± 1")[$± 1$] exactly at #math.equation(block: false, alt: "k equals ± 1")[$k = ± 1$] — the red and blue roads. Different roads, different constants: the two-variable limit does not exist.
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Checking a limit along a few roads can only ever #emph[disprove] it: one disagreement kills the limit, but agreement along every straight line is still not proof — curved roads get a vote too. Treat every two-variable limit as a claim about all roads at once.
]