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I want to calculate the following limit

$$L=\lim_{r \to 0} \left( 1 + r \frac{El^{'}(r)}{El(r)} \right)$$

by letting Mathematica to know

$$\lim_{r \to 0} \frac{El^{'}(r)}{El(r)} = A $$

where $A$ is a constant. It is evident that we should have $L=1$.

I used the following

Limit[1 + (r Derivative[1][El][r])/El[r], r -> 0, 
Assumptions -> {Limit[Derivative[1][El][r]/El[r], r -> 0] == a}]

But it didn't work.

What should I do?

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    $\begingroup$ Maybe replace that quotient by (A+O[r])? $\endgroup$ Jul 5, 2016 at 16:02

2 Answers 2

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Assume analyticity :

Limit[1 + (r El'[r])/El[r], r -> 0, Analytic -> True]

(*  1  *)

Analytic->True assumes that generic functions (e.g., El[r] and El'[r] in this case) are analytic.

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  • $\begingroup$ Can you kindly add some explanations that what the option Analytic really does, in your answer? :) $\endgroup$ Jul 5, 2016 at 20:32
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    $\begingroup$ @H.R. - See edit. If you need any more you will need to talk to a mathematician. $\endgroup$
    – Bob Hanlon
    Jul 5, 2016 at 20:58
  • $\begingroup$ That seems fine. Thank you. So by turning this option on, Mathematica understands that $\frac{El^{'}(r)}{El(r)}$ is analytic and hence its limit equals the value of the function? right? $\endgroup$ Jul 5, 2016 at 21:00
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    $\begingroup$ @H.R. - "... you will need to talk to a mathematician" $\endgroup$
    – Bob Hanlon
    Jul 5, 2016 at 21:08
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In addition to what was suggested by @BobHanlon, one might just replace the quotient of El'[r])/El[r] by the assumed value of A+O[r] to denote that it is the given constant A to first order.

In[646]:= Limit[1 + r (A + O[r]), r -> 0]

(* Out[646]= 1 *)
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