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# Taking a limit inside a function and graphing it

Posted 10 years ago
 Hello, thank you for your time.I am trying to graph the following function:f[x_] := 1/(Exp[(x-u)/(k*T)]+1);Plot[f[x_],{x,0,1}]I tried doing the above while setting T=0, but quickly learned that wouldn't work.   So I tried the following:f[x_] := 1/(Limit[Exp[(x-u)/(k*T)],T->0]+1);Then I tried to use the same plot.  This gave me no errors but now unfortunately nothing is showing up on my graph.  It should look like a step-function, where f(x) = 1 until x=u, then at that point f(x)=0 the rest of the way.Any help is appreciated, thank you!
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Posted 10 years ago
 I actually got it working. I had to make f a function of x and T, like this:f[x_,T_]...Manipulate[Plot[f(x,T)......]
Posted 10 years ago
 AHA got it! Its because I was using Plot[f[x_]]... I should have been using Plot[f]Thanks everyone!I do have one more question.  What If I want to useManipulate[Plot[f(x)],{x,0,10},{T,0,300}]Will this work?  Because I am trying just that and it doesnt seem to be working
Posted 10 years ago
 Bill, thank you so much for the sample!  Maybe I am doing something wrong, but it is still not working.  Here is the entirety of the code:k=8.617*10^(-5);In:=u=5.49;x=0;f(x_):=1/(Limit[Exp[(x-u)/(k*T)],T->0]+1);plot[f[x_],{x,0,10},AxesOrigin->{0,-.1}] When I compile, nothing shows up on the graph
Posted 10 years ago
 In:= u = 5;k = 10^-5;f[x_] := 1/(Limit[Exp[(x - u)/(k*T)], T -> 0] + 1);Plot[f[x], {x, 0, 10}, AxesOrigin -> {0, -.1}]Out= ...PlotSnipped...
Posted 10 years ago
 Hi thanks for the response.  I do have values for these constants. u=5, k=10^(-5), but T is approaching zero.
Posted 10 years ago
 Need numerical values for u, k, T to plot a function of x. Possibly introduce ranges of values using Manipulate, In Plot, should have f not f[x_]. Sometimes missing plot region can be revealed by adding PlotRange->All.