# Using Limit with the Option Assumptions

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?

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

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.

• Can you kindly add some explanations that what the option Analytic really does, in your answer? :) Jul 5, 2016 at 20:32
• @H.R. - See edit. If you need any more you will need to talk to a mathematician. Jul 5, 2016 at 20:58
• 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? Jul 5, 2016 at 21:00
• @H.R. - "... you will need to talk to a mathematician" Jul 5, 2016 at 21:08

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 *)