Recall: [math]\lim_{x\to a} f(x)=L[/math] if [color=#c51414]for every number [math]\epsilon >0[/math][/color], [color=#198f88]there exists a corresponding number [math]\delta>0[/math] such that for all [math]x[/math], [math]0<|x-a|<\delta[/math] [math]\Rightarrow[/math] [math]|f(x)-L|<\epsilon[/math][/color].[br]Illustration: [br] 1. Check the box for [math]\varepsilon[/math] - challenge for [math]H[/math][br] 2. Move [math]H[/math] on the [math]y[/math]-axis[br] 3. Use the vertical slider to give [color=#c51414]different [math]\varepsilon[/math] - challenge[/color][br] 4. Use the horizontal slider to [color=#198f88]adjust [math]\delta[/math] so that the [math]\varepsilon[/math] - challenge is passed[/color].[br]Observation: For choices [math]H=0.5,2[/math] or [math]3[/math], they fail the [math]\varepsilon[/math] - challenge.

Aim: To check that [math]L=1.69[/math] is the limit of [math]f\left(x\right)[/math] at [math]x=a[/math]. [br]  (i) Uncheck the box for [math]\varepsilon[/math] - challenge for [math]H[/math][br]  (ii) Check the box for [math]\varepsilon[/math] - challenge for [math]L[/math][br]  (iii) Repeat Steps 3 & 4[br]Observation: For the choice of [math]L=1.69[/math], the [math]\varepsilon[/math] - challenge is passed! So [math]\lim_{x\to a} f(x)=1.69[/math]