MapleLimitExNonLin.mws

Non-Linear Limit Example

limit(x^2+4,x = 5)

This example looks graphically at satisfying the epsilon-delta definition of limit with an epsilon value of 0.01 in considering limit(x^2+4,x = 5) = 29 .

>    with(plots):

>    f:=x->x^2+4;

f := x -> x^2+4

The first value looked at is delta = epsilon/10 .

>    left:=plot([4.999,t,t=28.99..29.01],thickness=2,color=blue,labels=[x,y]):

>    right:=plot([5.001,t,t=28.99..29.01],thickness=2,color=blue):

>    bottom:=plot([t,28.99,t=4.999..5.001],thickness=2,color=blue):

>    top:=plot([t,29.01,t=4.999..5.001],thickness=2,color=blue):

>    graph:=plot([t,t^2+4,t=4.999..5.001],thickness=2):

>    display(left,right,bottom,top,graph);

[Maple Plot]

>    f(4.999);

28.990001

>    f(5.001);

29.010001

Although the computations above clearly indicate the epsilon-delta definition of limit is not satisfied in this case, I will still try a little manual zooming.

>    rightZoom:=plot([5.001,t,t=29.009..29.01],thickness=2,color=blue,labels=[x,y]):

>    topZoom:=plot([t,29.01,t=5.0009..5.001],thickness=2,color=blue):

>    graphZoom:=plot([t,t^2+4,t=5.0009..5.001],thickness=2):

>    display(rightZoom,topZoom,graphZoom);

[Maple Plot]

Visually we cannot determine much from these pictures but we can still see from the computation below that the delta value is too large at the right hand side of the interval.

>    f(5.001);

29.010001

Thus time we are looking at delta = epsilon/11 .

>    left2:=plot([5-0.01/11,t,t=28.99..29.01],thickness=2,color=blue,labels=[x,y]):

>    right2:=plot([5+0.01/11,t,t=28.99..29.01],thickness=2,color=blue):

>    bottom2:=plot([t,28.99,t=5-0.01/11..5+0.01/11],thickness=2,color=blue):

>    top2:=plot([t,29.01,t=5-0.01/11..5+0.01/11],thickness=2,color=blue):

>    graph2:=plot([t,t^2+4,t=5-0.01/11..5+0.01/11],thickness=2):

>    display(left2,right2,bottom2,top2,graph2);

[Maple Plot]

This time we can visually observe that the epsilon-delta definition of limit is satisfied for an epsilon value of 0.01.  Since the function is an increasing function over the delta neighborhood of 5 being investigated we can also determine that the epsilon-delta definition of limit is satisfied based on the computations below.

>    f(5-.01/11);

28.99090992

>    f(5+.01/11);

29.00909174

This time we will be looking at delta = -5+sqrt(25+epsilon) .

>    left3:=plot([5-(-5+sqrt(25+0.01)),t,t=28.99..29.01],thickness=2,color=blue,labels=[x,y]):

>    right3:=plot([5+(-5+sqrt(25+0.01)),t,t=28.99..29.01],thickness=2,color=blue):

>    bottom3:=plot([t,28.99,t=5-(-5+sqrt(25+0.01))..5+(-5+sqrt(25+0.01))],thickness=2,color=blue):

>    top3:=plot([t,29.01,t=5-(-5+sqrt(25+0.01))..5+(-5+sqrt(25+0.01))],thickness=2,color=blue):

>    graph3:=plot([t,t^2+4,t=5-(-5+sqrt(25+0.01))..5+(-5+sqrt(25+0.01))],thickness=2):

>    display(left3,right3,bottom3,top3,graph3);

[Maple Plot]

This one is visually inconclusive too but in this case the value chosen for delta does satisfy the epsilon-delta definition of limit.  In fact this delta value is the largest value of delta we could have chosen if the epsilon value is to be 0.01.  Since we are looking at an increasing function we can see that the function value corresponding to an x value just less than 5 plus this delta value will be less than 29.01 as required.

>    f(5-(-5+sqrt(25+.01)));

28.99000200

>    f(5+(-5+sqrt(25+.01)));

29.01000000

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