Simple question about symbolic limits

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Tristan Potter
Tristan Potter 2020-6-9
编辑： Tristan Potter 2020-6-13
Hello,
I am checking results derived by hand in a MATLAB (2019a) live script and have encountered the following problem (MWE below): When I try to take the limit of the expression as (symbolic) variable l approaches 0, restricting , MATLAB cannot find the limit I obtain by hand. However, when I choose arbitrary values for instead of using an assumption, I get the same limit I derived by hand (which is independent of c). I suspect there is an obvious explanation for this that I am overlooking.
MWE:
%Declare symbolic variables:
syms l real;
syms t real; assumeAlso(t>1);
syms a real; assumeAlso(0<a<1);
syms p real; assumeAlso(0<p<1);
syms e real; assumeAlso(e>0);
syms c real; assumeAlso(c>0);
syms w real; assumeAlso(0<w<1);
%Functions:
phi = e/(1+c)*l^(1+c);
z = t/l*(1-exp(-l/t));
n = p*(1+c)/(p*(1+c)+z*l/phi*w);
pn = t*p*(1-n)*exp(-l/t)/n*(1-exp(-1*(t*p*(1-n)*exp(-l/t)/(a*n))^(-1)));
%Limits:
limit(pn,l,0,'right')
MATLAB cannot reduce this limit. However, when I instead impose, e.g., , via
syms l real;
syms t real; assumeAlso(t>1);
syms a real; assumeAlso(0<a<1);
syms p real; assumeAlso(0<p<1);
syms e real; assumeAlso(e>0);
syms c real; assumeAlso(c>0);
syms w real; assumeAlso(0<w<1);
%Functions:
phi = e/(1+0.1)*l^(1+0.1);
z = t/l*(1-exp(-l/t));
n = p*(1+0.1)/(p*(1+0.1)+z*l/phi*w);
pn = t*p*(1-n)*exp(-l/t)/n*(1-exp(-1*(t*p*(1-n)*exp(-l/t)/(a*n))^(-1)));
%Limits:
limit(pn,l,0,'right')
I get the result I derived by hand. The same holds for seemingly all other positive values of c. Any idea what I'm overlooking?
Thanks in advance.

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回答（1 个）

Ayush Gupta
Ayush Gupta 2020-6-12
The limit specified is also dependent on c if c does not have small values. For small values of c, a simplified expression is yielded. Refer to the following link to know more about limit:
https://www.mathworks.com/help/symbolic/limit.html
  1 个评论
Tristan Potter
Tristan Potter 2020-6-13
编辑：Tristan Potter 2020-6-13
Thanks for your response. However, I don't see that this resolves the issue for several reasons:
1) I don't see the dependence on c in a manual derivation of the limit via L'Hopital's rule (the limit I obtain manually corresponds to what MATLAB gives when I use a particular value of c, so I don't think I'm making a mistake in the manual derivation).
2) If I substitute the expression for n into the expression for above, I obtain:
pn = (t^2*(w-b)/e*exp(-l/t)*(1-exp(-l/t))/l^(c+1)*exp(-a/t*exp(-l*z)/exp(-l/t))*(1-exp(-a*(t^2*(w-b)/e*exp(-l/t)*(1-exp(-l/t))/l^(c+1))^(-1)));
When I evaluate this limit, seemingly any value of c gives the same limit that I get when I do the derivation manually. For example, I can choose and I get the same answer.
3) Finally, I tried writing where and applying L'Hopital's rule manually by taking the derivative of the numerator and denominator. I then used MATLAB to take the limit of the resulting ratio of the derivatives. When I simplify the ratio of derivatives and take the limit in MATLAB, I get the same result that I get manually. When I don't simplify, the problem remains:
n0 = 1-exp(-a*(t^2*(w-b)/e*exp(-l/t)*(1-exp(-l/t))/l^(c+1))^(-1));
d0 = 1/(t^2*(w-b)/e*exp(-l/t)*(1-exp(-l/t))/l^(c+1));
n1 = diff(n0,l);
d1 = diff(d0,l);
limit(n1/d1,l,0,'right')
limit(simplify(n1/d1),l,0,'right')
Am I missing something?

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