Mistakes

THE FOLLOWING IS FALSE:
$$\frac{c}{a+b}=\frac{c}{a}+\frac{c}{b}$$

FAKE EXAMPLE OF THIS ERROR BEING MADE:
Student writes:
$$\frac{x^2}{1+x^2}=\frac{x^2}{1\vphantom{x^2}}+\frac{x^2}{x^2}$$

TRY IT WITH NUMBERS:
$$\frac{2}{1+1}=1$$
$$\frac{2}{1}+\frac{2}{1}=4$$

For what real values of $a$ and $b$ and $c$ do we have equality?
We note that $a$ and $b$ are not equal to $0$. If $c=0$ it is true. If $c\neq 0$ then divide both sides by $c$ and we get $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$. Then let's combine the terms on the right-hand side of the equation. We get $\frac{a+b}{ab}$. Cross multiplying we obtain $ab=(a+b)^2=a^2+2ab+b^2$. Then $0=a^2+ab+b^2$. This is a quadratic equation. Let's solve for $a$ in terms of $b$. The discriminant is $b^2-4b^2=-3b^2<0$. There are no real values of $a$ and $b$ for which the equation is true. So basically, this equation is not just false some of the times, it's false most of the time. THE FOLLOWING IS INCOMPLETE: When making a $u$-substitution or trig substitution: $$u=f(x)$$ THE FOLLOWING IS INCORRECT:
When making a $u$-substitution or trig substitution:
$$u=f(x)$$
$$du=f^\prime(x)$$

THE FOLLOWING IS CLEAR AND CORRECT:
When making a $u$-substitution or trig substitution:
$$u=f(x)$$
$$du=f^\prime(x) dx$$

Find the difference.

POSSIBLE CONSEQUENCE OF THE FIRST ERROR (INCOMPLETE):
Student writes:
trig substitution
$x=\tan\theta$
$$\int \frac{1}{x^2\sqrt{1+x^2}}dx=\int \frac{1}{\tan^2\theta\sqrt{1+\tan^2\theta}}$$

POSSIBLE CONSEQUENCE OF THE SECOND ERROR (INCORRECT):
Student writes:
trig substitution
$x=\tan\theta$
$dx=\sec^2\theta$
$$\int \frac{1}{x^2\sqrt{1+x^2}}dx=\int \frac{\sec^2\theta}{\tan^2\theta\sqrt{1+\tan^2\theta}}$$

Note: Typically the student committing this mistake will still get the answer correct, but lack a sense of proper notation and/or presentation. Making the second mistake a milder one than the first, but it has been committed enough times that it gets on my nerves.

CONSEQUENCE OF DOING IT CORRECTLY:
Student writes:
trig substitution
$x=\tan\theta$
$dx=\sec^2\theta d\theta$
$$\int \frac{1}{x^2\sqrt{1+x^2}}dx=\int \frac{\sec^2\theta}{\tan^2\theta\sqrt{1+\tan^2\theta}}d\theta$$


THE FOLLOWING IS FALSE:
$$\infty - \infty = 0 $$

FAKE EXAMPLE OF THIS ERROR BEING MADE:
Student writes:
$$\lim_{x\to \infty} \ln(2x+1)-\ln(x+2)=\infty -\infty =0$$

HOW TO DEAL WITH THIS PARTICULAR EXAMPLE:
$$\lim_{x\to \infty} \ln\left(\frac{2x+1}{x+2}\right)=\ln{2}$$


SIMILAR PROBLEMS WITH OTHER INDETERMINATE FORMS:
$1^\infty=1$

FAKE EXAMPLE OF THIS ERROR BEING MADE:
Student writes:
$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=1$ because $\lim_{n\to\infty} \left(1+\frac{1}{n}\right) =1$.

Note that the value of this limit is actually the constant $e$.


THE FOLLOWING IS FALSE:
The student assumes one part of the expression goes to infinity implies the entire expression diverges.

FAKE EXAMPLE OF THIS ERROR BEING MADE:
Student writes:
$$\lim_{x\to \infty} \ln(2x+1)-\ln(x+2)$$
$\lim_{x\to\infty} \ln(2x+1)=\infty$ and so the previous expression diverges.

When can the conclusion be made?
If all the remaining portion of the expression has a finite limit.


THE FOLLOWING IS FALSE:
Every series with a $(-1)^n$ magically becomes an alternating series.

FAKE EXAMPLE OF THIS ERROR BEING MADE:
Student writes:
$$\sum _{n=0}^\infty (-1)^n \frac{5^n \left(-\frac{2}{5}\right)^n}{2^n(n+1)}$$ converges by the alternating series test.

What's wrong with this?
If we actually multiply everything out, we get $$\sum_{n=0}^\infty \frac{1}{n+1}$$ which diverges by comparing it with the harmonic series or by applying the integral test.

On that note, I have some other mistakes that come up:
$\ln(a+b)=\ln(a)+\ln(b)$
Interpreting notation.