# Common Math Mistakes (by level)

## Common Mistakes

Common Mistakes

A big chunk of credit can be saved in math courses by avoiding what we math professors see as common mistakes. Each course in the math sequence has its own set of mistakes that students often make. The purpose of this page is to showcase these mistakes and to discuss why they are wrong and how to avoid them!

# Simplification Errors

Very common

###### Deduction:

These concepts are core to the rest of the math sequence. Making these errors in arithmetic is significant; however, deductions for these errors range by instructor for more advanced courses.

###### Issue:

Simplifying simple arithmetic expressions is a common task in any mathematics course. It not only allows you to reduce the complexity of a mathematical statement, it also makes additional computations much easier. An inability to simplify arithmetic expressions is a key indicator that a student is suffering from a severe lack of core understanding of the underlying mathematics in a course.

###### Example(s)

Either a student is directly asked to simplify any of the following arithmetic expressions or these expressions occur within a larger, more complex problem.

Unsimplified Expression Incorrect Simplification Correct Simplification
$$1 + 8$$ $$1 + 8$$ $$9$$
$$\frac{2}{10}$$ $$\frac{2}{10}$$ $$\frac{1}{5}$$
$$3 \cdot 8$$ $$3 \cdot 8$$ $$24$$
$$\frac{1}{4} + \frac{1}{3}$$ $$\frac{1 + 1}{4 + 3}$$ $$\frac{7}{12}$$
$$\sqrt{9}$$ $$\sqrt{9}$$ $$3$$
$$2^{5}$$ $$2^{5}$$ or $$2 \cdot 5 = 10$$ $$32$$
###### Remedy:

The best way to not get deductions for simplification mistakes is to constantly remind yourself of the following advice.

If you have learned a skill, you are expected to use that skill!

# Sloppy Graphs

Somewhat common

###### Deduction (instructor-dependent):

Major deductions if you were specifically asked to graph something within an exercise, but likely no deductions if you are only using the graph as a quick glimpse of a feature to a problem

###### Issue:

A major prerequisite to calculus is the ability to carefully construct a complete graph of a given function. At this point in your mathematical career, there are no excuses for sloppy graphs.  Point-plotting to get the graph of a function should be virtually unheard of. Moreover, when you are graphing shapes of the base graphs (parabolas, sine waves, logarithmic curves, etc.), the curves should be just that - curves! A sine wave that is a bunch of lines connecting maximums and minimums is worth nothing in calculus. It demonstrates to your instructor that your knowledge of these graphs is weak at best.

###### Remedy:

Your instructor will emulate what they expect from you. Be sure to pay attention to this! If you have weak graphing skills, then I recommend you look at your old precalculus, trigonometry, and algebra textbooks. We hope to have some videos up soon to help review this content; however, being weak in graphing usually is a sign of weak skills in many other areas.

# Not Justifying Limits (C1-JL)

###### Frequency:

Very common at the beginning of a Calculus I course

Full deductions

###### Issue:

In Calculus I, if asked to evaluate a limit, just writing an answer is not good enough.  The reason this is bad is because many limits tend to either $$\infty$$, $$-\infty$$, or they DNE. Just writing these answers starts to look like guessing. This is a calculus course and your logic must be justified.

###### Example(s):

A student is asked to evaluate $\lim_{x \to \pi^{-}}{\cot{(x)}}$ and responds only with $$-\infty$$.

###### Remedy:

You must justify your reasoning with logic.  For example, with the given example, you can justify this limit in one of two ways. Either you can draw a clean graph of the function and note its behavior as $$x$$ approaches the vertical asymptote at $$\pi$$ from the left, or you can rewrite the cotangent function and evaluate as follows: $\lim_{x \to \pi^{-}}{\cot{(x)}} = \lim_{x \to \pi^{-}}{\frac{\cos{(x)}}{\sin{(x)}}}.$You can then note that the numerator approaches $$-1$$ while the denominator approaches $$0^{+}$$. Thus, the limit must be $$-\infty$$.

# Distributing Unnecessarily (C-DU)

Uncommon

###### Deduction:

Depends on whether or not the distribution leads to a false statement, but it can lead to major deductions

###### Issue:

You are taught very quickly in calculus that a sum or difference (often occurring in a fraction) involving at least one radical term can be "cleaned up" by multiplying both numerator and denominator by the conjugate of that sum or difference. Distributing the conjugate factors is fine, but it is unnecessary, and often quite bad, to distribute the factors that are not conjugate. To illustrate the issue, see the example below.

###### Example(s)

Evaluate $\lim_{h \to 0}{\frac{\sqrt{x + h} - \sqrt{x}}{h\sqrt{x}}}$A student who sees this would appropriately multiply both numerator and denominator by the conjugate of the numerator, $$\sqrt{x + h} + \sqrt{x}$$, to eventually get $\lim_{h \to 0}{\frac{1}{\sqrt{x} (\sqrt{x + h} + \sqrt{x})}}.$However, multiplying out that denominator would lead the student to believe that this is the same as $\lim_{h \to 0}{\frac{1}{\sqrt{x^2 + hx} + \sqrt{x^2}}}.$This is not true! The original problem included an implied domain restriction that $$x > 0$$. This most recent answer has lost that restriction and has now (possibly) convinced an unwary reader that $$x$$ can be anything other than $$0$$.

This level of subtlety is caught only by the most advanced students in a calculus course and, therefore, it is advisable to learn how to avoid even remotely running into this case. I do not mean that you are not responsible for understanding why this case is an issue, but I would rather you not lose precious credit by introducing this subtlety into your computations.

###### Remedy:

While there are many other, less subtle, issues with distributing unnecessarily in calculus, it suffices to say that the best advice is to only distribute conjugate multiples or when the distribution is meant to clear complex fractions.

# Not Understanding Absolute Values (C-AV)

Common

Full deductions

###### Issue:

Most students think of the absolute value as a function that makes something positive. While this is true, the how it makes something positive is incredibly important in calculus.

###### Example(s)

Consider the limit $\lim_{x \to -2}{\frac{|x+2|}{x+2}}.$The wrong thing to do here would be to just "cancel" the $$x + 2$$ so you get the limit of $$1$$. You can't do this because $$|x+2|$$ is not equivalent to $$x + 2$$. In fact, the definition of the absolute value shows that$\lvert x + 2 \rvert = \begin{cases} x + 2 & \text{if } x + 2 ≥ 0 \Rightarrow x ≥ -2\\ -(x + 2) & \text{if } x + 2 < 0 \Rightarrow x < -2\end{cases}$Our limit is concerned with $$x \to -2$$ and our absolute value definition shows us that the value of $$|x + 2|$$ is different on either side of $$-2$$. Hence, we must tackle this limit from the left- and the right-hand sides separately.$\lim_{x \to -2^{-}}{\frac{|x+2|}{x+2}} = \lim_{x \to -2^{-}}{\frac{-(x + 2)}{x+2}} = -1.$This is because $$x \to -2^{-}$$ implies $$x < -2$$. Therefore, we use $$|x + 2| = - (x + 2)$$. On the other hand, $\lim_{x \to -2^{+}}{\frac{|x+2|}{x+2}} = \lim_{x \to -2^{+}}{\frac{x + 2}{x+2}} = 1.$Again, this is because $$x \to -2^{+}$$ implies $$x > -2$$. Therefore, we use $$|x + 2| = x + 2$$. Since the right- and left-hand limits are not equivalent in this case, the overall limit does not exist (DNE).

Let's change the limit slightly to see an interesting side case. Consider the limit $\lim_{x \to -2}{\frac{2 - |x|}{x+2}}.$Again, we use the definition of the absolute value$\lvert x \rvert = \begin{cases} x & \text{if } x ≥ 0 \\ -x & \text{if } x < 0 \end{cases}$However, notice that because $$x \to -2$$, $$x$$ is always less than $$0$$ (as long as it is close enough to $$-2$$). Therefore, we can exchange $$|x|$$ with $$-x$$ to get $\lim_{x \to -2}{\frac{2 - (-x)}{x+2}} = \lim_{x \to -2}{\frac{2 +x}{x+2}} = 1.$

###### Remedy:

For computational purposes, the real definition of the absolute value function is key! $\lvert f(x) \rvert = \begin{cases} f(x) & \text{if } f(x) ≥ 0\\ -f(x) & \text{if } f(x) < 0 \end{cases}$Did you notice that I used $$f(x)$$ instead of $$x$$? This is on purpose! A lot of students make the mistakes listed in the Example(s) section above without understanding that we are often not concerned about the sign of $$x$$, but rather the sign of $$f(x)$$.

To compound matters a bit, just relying on using the definition of the absolute value is not enough. You must be aware of when the definition is even necessary.

# Not Using Limits to Evaluate Limits (C-UL)

Common

###### Deduction (instructor-dependent):

Simpson: A warning the first time, but major-to-total deductions thereafter

###### Issue:

The entire point of a limit is that the independent variable is getting infinitesimally close to some value without ever attaining that value. With this concept comes much power in the way of computation (see the example below).

###### Example(s)

Suppose you are asked to evaluate $\lim_{x \to -2}{\frac{x^2 + 5x + 6}{x + 2}}$Many students decide to say, "Well, we know that $\frac{x^2 + 5x + 6}{x + 2} = \frac{(x + 2)(x + 3)}{x + 2} = x + 3$However, this is not true! This is akin to someone telling you that you can walk across an active minefield... without telling you to be sure and hop over a small bit of it about six feet away.

I get why it's so attractive to students to just do the algebra first without the limits, but... sorry folks, that algebra (and the lack of calculus) is just plain wrong. The actual work is$\lim_{x \to -2}{\frac{x^2 + 5x + 6}{x + 2}} = \lim_{x \to -2}{\frac{(x+2)(x+3)}{x+2}} = \lim_{x \to -2}{(x+3)} = 1$The only reason those factors of $$(x + 2)$$ can divide out is because we know that $$x$$ will never be $$-2$$!!!

###### Remedy:

Use limits when working with limit problems!