Using MAPLE to find a value of A

Question

I am trying to utilize Maple to find a value of A that makes $\lim\limits_{h \to 0} \frac{A^h-1}{h} = 1$

I started with $h = 1$ then I decrease $h$ by a factor of 0.1. I tried repeating the process until four or five decimal digits remained the same. Please help. Thanks so much.

Answer 1

This looks like coursework (which is fine), but you haven't really indicated how much (or less) you are expected to get Maple to do here. It can be used to solve the whole thing exactly (symbolically). Or it can be used to solve the whole thing approximately (ie. numerically, in floating-point). Or it can be used to do a mix of those.

One step is taking the limit, and the other is root-finding. (It's even conceivable that you are expected to code up your own numeric root-finder, but I won't show that unless you ask.)

restart;
eq := limit((A^h-1)/h,h=0)=1;

                           eq := ln(A) = 1

ans := solve(eq, A);

                            ans := exp(1)

# The previous step produced exp(1), which Maple
# can knows how to approximate efficiently.
evalf(ans);

                             2.718281828

restart;
eq := limit((A^h-1)/h,h=0)=1;

                           eq := ln(A) = 1

# Now solve this intermediate equation purely
# numerically, which is less efficient that the
# above.
ans := fsolve(eq, A);

                         ans := 2.718281828

restart;
# A purely numeric approach using the original
# equation and the inert Limit so that numeric limits
# are computed. (Of course this compares
# very poorly with the above, in terms of efficiency).
fsolve(Limit((A^h-1)/h,h=0)=1,A=1..3);

                             2.718281828

Answer 2

You can simply use the solve function:

solve(limit((A^h-1)/h,h=0)=1,A);

Using limit((A^h-1)/h,h=0) one computes the limit $$\lim\limits_{h\to 0}\frac{A^h-1}{h}.$$

Then using solve() we solve for the value of $A$ such that the limit equals $1$.

Answer 3

You can solve it without using Maple, it is a simple problem : $$\lim\limits_{h\rightarrow0}\frac{A^h-1}{h}=\lim\limits_{h\rightarrow0}\frac{A^h-A^0}{h-0}=(x\mapsto A^x)'(0)=\ln(A)A^0=\ln(A)=1\Rightarrow A=e.$$