The first sentence of the question is "A frequent issue with students is their uncritical over-reliance on computation devices," hence my answer is in the spirit of presenting something where the result from the calculator is useless, even if it is correct in some technical sense.

Calculators / graphing utilities are terrible at graphing anything which has a lot of variation in scale. For example, in a precalculus lecture last week, I had students graph the rational function

$$x \mapsto -\frac{(x^2-1)(x^2+4x+4)^3(x+2)}{(x^2-4)(x-3)^4(x+5)^3}. $$

Naively plugging this into GeoGebra gives something similar to

From this picture, many of the important features of the graph are missing. For example, it is hard to see that there are two zeros in $(-5,2)$, the "hole" in the graph at $x=-2$ is not apparent (and if you click on the graph, GeoGebra will note that there is a removable discontinuity, but there is numerical error in the reported location of the hole), the asymptotic behaviour of the function as as $x \to \pm \infty$ isn't very clear, and it is impossible to see what is happening on $(2,3)$. A lot of useful information is lost.

On the other hand, my hand-drawn graph is

The vertical and horizontal scales in my picture are total nonsense, but the asymptotes are shown, the zeros are located, the removable discontinuity is marked, and the general behaviour of the function is a bit easier to see. This picture incorporates information about singularities, zeros, and limiting behaviour—in a calculus class, we might also locate local extreme values, and incorporate information about monotonicity and concavity into the picture.

A simple question about this function is "What is the natural domain?" A student who plugs the function into a graphing utility and tries to read off the graph will get it wrong. They will almost certainly miss the removable discontinuity entirely, and are very likely to exclude an interval like $[2,6]$ from the domain. Indeed, I often catch students cheating because they make exactly this kind of mistake.

The last time I taught introductory numerical analysis, my first slide was this print from excel:

If you want another example:

These calculations are clearly easy and wrong, as required.

Remark 1: Honestly, I don't know if these errors are a concern given that even NASA only needs 15 decimal places.

Remark 2: The presented errors result from the representation of repeating binary numbers.

Evaluate $L=\lim\limits_{x\to 0^+}(\sin x)(\ln x)^{100}$.

Letting $f(x)=(\sin x)(\ln x)^{100}$, students might use their calculators to find that:

$f(0.1)\approx 1.66\times 10^{35}$
$f(0.01)\approx 2.11\times 10^{64}$
$f(0.00001)\approx 1.31\times 10^{101}$

and then conclude that $L=\infty$ (does not exist).

But actually $L=0$.

$\begin{align} L&=\lim\limits_{x\to 0^+}(\sin x)(\ln x)^{100}\\ &=\lim\limits_{x\to 0^+}\frac{(\ln x)^{100}}{(\sin x)^{-1}}\\ &=\lim\limits_{x\to 0^+}\frac{100(\ln x)^{99}\left(\frac{1}{x}\right)}{-(\sin x)^{-2}\cos x}\space\text{L'Hopital's rule}\\ &=-100\left(\lim\limits_{x\to 0^+}\frac{\tan x}{x}\right)\left(\lim\limits_{x\to 0^+}(\sin x)(\ln x)^{99}\right)\\ &=\dots\\ &=100!\lim\limits_{x\to 0^+}(\sin x)(\ln x)^0\\ &=0\\ \end{align}$

The easiest example off the top of my head is evaluating $(9^n)^{1/n}$ for a ridiculously large value of $n,$ large enough to have the calculator throw an overflow error. Something like $(9^{9^9})^{(1/9^9)}$ usually does it.

But honestly, if the heart of this question is just trying to come up with a good reason why students shouldn't over-rely on calculators, I don't think this is the best direction to look.

A better reason would be that if you over-rely on a calculator, you won't develop automaticity on the skills that you're offloading to the calculator, and that will prevent you from building on top of those skills.

For instance, if someone always uses a calculator and doesn't build automaticity on their arithmetic skills, then they're going to struggle in algebra. Even if they manage get through algebra, they won't have developed automaticity on their algebra skills, and consequently they're going to struggle even more in the next math course they take.

The easiest of all:$$\frac{1}{3} \cdot 3$$ :-)

You can also go for large squares, something like:

$$\frac{123456789^2 - 123456788^2}{123456789 + 123456788}$$

=> for a human, this is just $\frac{a^2-b^2}{a+b} = a - b$

Ask a TI-89 to evaluate $\sin(n \pi)$ for different integers $n$, and you'll start out fine (when $n = 0, 1, 2, 3$), but starting with $n = 4$, you'll get values like $-2 E -13$.

Or there's $8^{8^8}$ which has the output $\infty$ on that calculator.

Combinatorics might be a good place to look.

A very basic example: given 200 items, how many ways are there to pick 1 item?

Using the permutation formula involves calculating $200!/199!$, which not even Google's calculator can manage.

But it's very easy to do by hand:

$200!/199! = (200)(199!/199!) = 200$

In a rather different vein, calculators can be correct in a numerical sense but miss an optimal solution method. In this answer two such cases are described, and I refer the reader to the above link for details:

• The quartic equation $x^4+x^3+x^2-x+1=0$ can be solved by setting up a factorization $(x^2+ax-1)(x^2+\overline{a}x-1)$ and solving quadratic equations. Wolfram Alpha recognizes such a solution, but a TI-89 calculator gives only numerical answers.

• The quintic equation $x^5+10x^3-5=0$ is provably soluble by radicals and gives a relatively simple form for the roots, but even Wolfram Alpha misses this and gives only numerical answers.

Occasionally I ask my students true-false questions of the following form:

• T/F? $1/3 = 0.333333333$
• T/F? $\pi = 3.141592654$

They usually get these right as a group, but I think it's good to check in on it.

Many cheap calculators ignore operator precedence. Type 1+2*3 on a good one and you'll get 7, do it on a cheap one and you get 9.

And then there's the anecdote a math teacher friend told me where it's unclear if the square of -5 is -25 or 25.

Continuing in the vein of "The calculator is right, but the interpretation by the user is wrong" I add a class of problems of the form

$$\lim_{x \to \infty} \frac{p(\log(x))}{q(\log(x))}$$

where $p$ and $q$ are polynomials. While another user has given a specific answer in this vein that uses L'Hopital's rule, I think adding a general class of functions might be of some use as well. This class of limits are also readily evaluated using elementary techniques (e.g. without funny business like L'Hopital's rule), and so they can be introduced early in the calculus course (around when you need to get buy-in from students about this very issue).

One such example is evaluating the limit, $\lim_{x\to \infty}\frac{\log(x)}{10+\log(x)} $

The answer is, of course, $1$. However, students may be inclined to estimate the limit numerically or visually. A graph of the function will not readily yield this limit. In fact, one must enter values of at least $10^{40}$ to reach a function output of at least 0.8, values of at least $10^{90}$ to reach an output of at least .9, at which point the graph looks visually entirely flat; reaching what students might consider the "gold standard" of 0.99 requires inputs of at least $10^{990}$, whereas Desmos won't graph past the $10^{100}$ scale. Both entering values and looking naively at a graph of the function will lead to difficulty obtaining the correct result.

Computers are quite bad at drawing curves with self-intersections. Examples below are made with Maple 13 (a quite dated version).

The graph of $x^2+y^2=3xy$ looks particularly strange.

We can get a better figure by forcing the software to compute more values.

Even with $1000$ points, the graph may be confused with a hyperbola. It is actually two lines that intersect at the origin.

Finally, here is an example of a polar curve $r=\sqrt{\cos(2\theta)}$.

Some scientific calculators will round certain rational numbers to rational multiples of pi. For example, $\frac{11^6}{13} \approx \frac{156158413}{3600} \cdot \pi$. They may be assuming that the user is converting a rational multiple of degrees or gradians to radians.

Source: Stand-up Maths (youtube)

Here is a simple example. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

  $\color{green}{\Big( 99 · \big( 100·\sum_{k=0}^{10} (-\lfrac1{99})^k-99 \big)^\lfrac1{10} - 1 \Big)^\lfrac1{9} = 0}$.

Windows Calc ( 99*( 100*(1-1/99+1/99y2-1/99y3+1/99y4-1/99y5+1/99y6-1/99y7+1/99y8-1/99y9+1/99y10)-99 )y(1/10) - 1 )y(1/9) = gives 0.0117619348321432859010038254248.

My Casio scientific calculator does not have enough input space for the whole expression so I used it to first compute $\big( 100·\sum_{k=0}^{10} (-\lfrac1{99})^k-99 \big)$, which gives 0, and then used it to next compute $(99·0^\lfrac1{10}-1)^\lfrac19$, which gives -1.

Note that both of these results are as expected, because the expression is intentionally constructed to create and amplify rounding errors.

Since I was asked to post a comment of mine as an answer to my own question, here is a refined version:

Problem: Evaluate $$\text{log}_{10}(10^{\frac{1}{10^{17}}})^{-1}.$$

By hand, this easily evaluated to be $10^{17}$, but, for example, Python 3.10 complains about a "division by zero" error:

Input:

import math
math.log10(10 ** (1 / 10**17)) ** (-1)

Output:

ZeroDivisionError: 0.0 cannot be raised to a negative power

Wolfram Alpha still gets it right (and so will Python, when you change the datatype), but in that case, it suffices to replace 17 with something larger. (For 99999, Wolfram Alpha does not return an answer, and plugging in 999999999 makes Wolfram Alpha return "indeterminate".)

Another impressive example, found here in a worksheet called "rounding errors lab" https://cse.unl.edu/~scooper/securecoding/RoundingErrorsLab.pdf is that Google will compute 1-0.999998=0.00000199999.

An example discussed in math.se in the past is this: $-4^2$ is interpreted by all mathematicians as $-(4^2) = -16$. But some calculators (and even some elementary-school teachers) will interpret it as $(-4)^2 = 16$.

--added--
Of course I am talking about algebraic calculaors, where you can enter something like -4^2.

example
In Tiger Algebra, I typed -4^2

When I was looking for this example, I was pleased to find that many on-line algebraic calculators do this the other way. Are things improving?

Type the following into Wolfram Alpha:

integrate sin(pi x)/(x(1-x)) from x = 0 to x = 1

Half the time Wolfram Alpha gives a correct estimate and half the time it says the integral is $0$, even when it displays a valid graph showing the function is positive on $[0,1]$. If you try that and get a positive answer the first time, then input the request a second time and, in my experience, Wolfram Alpha will say the integral is $0$.

This glitch was brought to my attention several years ago and it's still there (I just checked).

Division by 0

This is pushing the definition of "easily be solved by hand" since the "solution" is to see that it's indeterminate.

Which calculators have this issue?

When you try 1/0 on the system calculator that comes with Windows XP you'll get "Infinity". This is wrong, however Windows 8 no longer has this problem and I doubt that your students will try to rely on Windows XP.

For a modern example https://www.wolframalpha.com/input?i=1%2F0 returns "complex infinity". This is intentional behavior, however if you are working strictly with real numbers then this isn't the right answer.

Why this answer matters

This answer is pretty lame but it has lead to confusion before. I met some students in high school and college who thought "Infinity is the correct answer to 1/0 but that's only for advanced math" however this isn't true. Even for surreal numbers (AFAIK) the result of 1/0 is undefined. I'm guessing this misconception came about because of the Windows XP calculator (and maybe some other old calculators with the same problem) so this is likely not an issue anymore.

Where this answer does still matter is for programming. The FPU and many programming languages intentionally have 1/0 = Infinity because that's the intended way to obtain an Infinity value from the FPU. Programmers who aren't aware of this may not realize that 1/0 doesn't result in NaN. Failure to account for division by 0 will lead to calculation bugs such as that in Windows XP.

Other programming issues

Everyone else has already commented on rounding errors and precision limitations. These are called underflow and overflow errors. There's also truncation and other issues. When programming math it's important that you understand the math!

A simple example, due to Velvel Kahan (UCB), is $$(((4/3)-1)*3)-1$$ is usually not zero (and should not be zero on IEEE-754 compliant floating-point systems; the HP 35s yields $-1{\rm e}{-}11$). Note parentheses. It normally yields the "machine epsilon", the smallest number when added to $1$ yields a sum different than $1$. This is the smallest difference between (binary64) floating-point numbers lying in the interval $[1,2)$.

This example does not work on the Apple iPhone (added: or Google and some others — see comment below), which has some strange numerics (I suspect "cosmetic rounding" but I cannot confirm). On the iPhone, compute $$(((258/257)-1)*257)-1$$ or $$((4/3)-1)-(1/3)$$ These do not result in machine epsilon, however.

For a TI-84, this variation returns the (negative of) the epsilon: $$(((100/3)-33)*3)-1 \mapsto -1{\rm e}{-}12$$

On the HP Prime Lite app (iOS), $$(((10/3)-3)*3)-1 \mapsto -1{\rm e}{-1}11$$

On the Google calculator, we have $$(((1000/3)-333)*3)-1 \mapsto 2^{-44} \approx -5.6843419{\rm e}{-}14 $$ which is, in Matlab code, eps(1000/3), the epsilon relative to 1000/3 in binary64 (double precision floats).

Similarly $0.3 - (3*0.1)$ is not zero on binary IEEE-754 compliant calculator/computer systems (again, it's zero on iPhones). It's not hard to come up with similar examples (e.g., $(23*0.1 - 20*0.1) - 3*0.1$).

On decimal calculators, the answer should be zero (no rounding error).

Why not problems like $(-1)^z=2$, or $i^z=3$? Even better if you ask for all solutions in their exact form. ($i$ in this case being defined as $\sqrt{-1}$)

For both of these examples:$$(-1)^z=2\\\text{We can use }(-1)=e^{2i\pi n}\text{ where }n\in\mathbb Z\\\text{So it becomes }e^{2i\pi nz}=2\\\implies2i\pi nz=\ln2\\\implies z=\frac{\ln2}{2i\pi n}=\frac{-i\ln2}{2\pi n}\quad n\in\mathbb Z$$And then$$i^z=3\\\text{We can use }i=e^{i\pi/2+2i\pi n}\text{ where }n\in\mathbb Z\\\text{So it becomes }e^{(i\pi/2+2i\pi n)z}=3\\\implies(i\pi/2+2i\pi n)z=\ln3\\\implies z=\frac{\ln3}{i\pi/2+2i\pi n}=\frac{2\ln3}{i(\pi+4\pi n)}\\=\frac{-2i\ln3}{\pi+4\pi n}\quad n\in\mathbb Z$$The reason I'm saying it's even better if you ask for all solutions in exact form is because even if their calculators can solve this in exact form, most likely it's just going to be one possible solution, rather than all possible solutions.

Here's an expression that evaluates to something very clearly inaccurate on most calculators:

$$\frac{1}{\left(\left(\frac{4}{3}\right)-1\right)\cdot 3-1 + 10^{-20}}$$

Most floating point calculators should give something like   -4503802460601329.0, but the correct answer is of course $10^{20}$, a factor of $10^5$ larger. The relatively small floating point errors add up.

Division 8:4:2.

I guess that most calculators will provide 1 or 4 as a result while in fact this is not defined as an arithmetic operation.

Here are two simple examples of computational limitations: (1) Calculator precision: In a common hand calculator, the fractions

a = 8,712,870/48,506,557 and b = 505,149/2,812,281

both display as 0.179622519. However, these are actually different numbers—the calculator's limited precision makes them appear identical. (2) GeoGebra error: In GeoGebra, the command Intersect(exp(-x) + x, x) returns the point (34.77474, 34.77474). However, the graphs of exp(-x) + x and the line y = x have no intersection points. This is a computational error since exp(-x) + x > x for all real x.

Most calculators completely ignore the concept of a multivalued function.

A very basic example:

>>> [4] [√]
2

What we got is just the principal square root. No calculator, save for a computer algebra system, reports the other root $-2$.

Notice how basic this example is: all the roots roots are real, and there are only two (i.e. a finite number) of them. There are more complex cases, such an $n$ roots of $\root n\of {x}$ (for $x \neq 0$; if we allowed the results to be complex), or infinitely many roots of functions with multiple branches, such as inverse trigonometric functions or a (complex) logarithm.

To demonstrate that to students, think of a math problem where using the principal value blindly (given by a calculator) leads to contradictory results.