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matlabsymbolic-math
Jerry
Jerry

Reputation: 4408

Matlab: Using sym in limit

Say I have the following symbolic function:

syms f(t)
f(t)=sin(t)/t

I want to get the limit using another symbolic function. I tried:

syms lim(x)
lim(x)=limit(f(t),t,x)

But when I tried to use lim(0) I got this error:

Error using symengine (line 59) Division by zero.

Can this be fixed?

Upvotes: 1

Views: 369

Answers (2)

TroyHaskin
TroyHaskin

Reputation: 8401

Matlab does not have delayed assignments as discussed here. Therefore, when lim is created, the call to limit is immediately evaluated with x replacing t:

>> syms t x f(t) lim(x)
>> f(t) = sin(t)/t
f(t) =
sin(t)/t

>> lim(x) = limit(f(t),t,x)
lim(x) =
sin(x)/x

And when you evaluate lim(0), you get sin(0)/0, which throws the error.

Upvotes: 1

Daniel
Daniel

Reputation: 36710

Take a look at lim(x). For some reason the limit is gone. I don't really understand what is going wrong there. If you use an anonymous function instead of a function handle, the evaluation of limit is postponed until x has a value and it works.

>> lim=@(x)limit(f(t),t,x)

lim = 

    @(x)limit(f(t),t,x)

>> lim(0)

ans =

1

Upvotes: 1

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