Can a calculator find any limit?
I am using CASIO fx-991ex and I noticed through my homework that when x approaches infinity or zero it's almost like using big or small numbers in calculator. For example $e^x$ while $x$ approaches $\infty$ is a very big number like using $e^9$ (+-) in the calculator, or $1/x$ while x approaches $0$ with $0.0001$ in the calculator (instead of $x$) it returns also a big number which I can assume it's $\infty$. So my question is when this thing doesnt work? I know that knowing how to find it out by myself is another important thing but im asking mainly for verifying purposes.
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$\begingroup$As a general principle it's fine to use a calculator as a tool to reinforce your intuition or guess an answer or confirm a theoretical calculation.
It's not a substitute for logical analysis.
There will be times when it "doesn't work" for several reasons. The calculator works with limited precision (just a finite number of decimal places) and finite range (there is a largest possible value). It may also fail when the limiting value is approached with oscillations rather than monotonically as in your two examples.
$\endgroup$ 1 $\begingroup$The Ackermann function is defined as:
$$A(0, n) = n+1$$$$A(m, 0) = A(m-1, 1)$$$$A(m, n) = A(m-1, A(m, n-1))$$
Define $A(n) = A(n, n)$.
The function $A$ grows extremely quickly. $A(3) = 61$. $A(4)$ is bigger than $10^{1355718576299609}$ - likely much bigger, but that is just the easiest lower bound estimate I got from Mathematica from the fact that computation overflows.
Define the inverse Ackermann function, $a(n)$, as the inverse of this function. So $a(61) = 3$. For any number your calculator can represent, the value of $a$ is less than $4$. Moreover, for any number Mathematica can represent on my machine, the value of $a$ is less than $4$. And yet the limit of $a(n)$ as $n \to \infty$ is $\infty$.
Or consider $\sum_{n=1}^{\infty} \frac{1}{n}$. This is well-known to be infinite (by the Cauchy condensation test, for example). And yet $\sum_{n=1}^{1,000,000} \frac{1}{n}$ is approximately $14.3927$. If you use your calculator to estimate the value of the limit, you will get it wrong.
$\endgroup$ 2 $\begingroup$The process that you are using there tends to work all the time. It's exactly what I did in highschool when I knew nothing about limits. Like taking $x\to\infty$ I just typed $x=999999$ into my calculator and it gave me the correct limit. This was especially applicable when we were looking at the logistic equation for population growth. It worked similarly with taking $x\to 0$ as you said.
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