### Examples of calculus problems

Here is a collection of examples of calculus questions with hints, answers, solutions, common mistakes and other comments. You can click on the corresponding items to show/hide their content.

Compute the limit. If the limit does not exist, determine with proof whether the trend is $\infty$, $-\infty$ or neither.

$\qquad\displaystyle\lim_{x\to 3}\frac{x-3}{\sqrt{3x}-3}$

**Answer:** 2

Rationalize the denominator. Namely, any difference of square roots $\sqrt{a}-\sqrt{b}$ for non-negative $a$ and $b$ can be rationilized by multiplying and dividing by the sum of these roots. Then

$\qquad\displaystyle\sqrt{a}-\sqrt{b}=\frac{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$.

For our denominator, take $a=3x$ and $b=9$.

$\qquad\displaystyle \lim_{x\to 3}\frac{x-3}{\sqrt{3x}-3}=\lim_{x\to 3}\left(\frac{x-3}{\sqrt{3x}-3}\cdot\frac{\sqrt{3x}+3}{\sqrt{3x}+3}\right)$

$\qquad\qquad\displaystyle =\lim_{x\to 3}\,\frac{(x-3)(\sqrt{3x}+3)}{3x-9}=\lim_{x\to 3}\,\frac{(x-3)(\sqrt{3x}+3)}{3(x-3)}$

$\qquad\qquad\displaystyle =\lim_{x\to 3}\,\frac{(x-3)(\sqrt{3x}+3)}{3x-9}=\lim_{x\to 3}\,\frac{(x-3)(\sqrt{3x}+3)}{3(x-3)}$

$\qquad\qquad\displaystyle =\lim_{x\to 3}\,\frac{\sqrt{3x}+3}{3}$

$\qquad\qquad\displaystyle =\frac{3+3}{3}=2$

- Improper use of limit notation. For example, not writing $\displaystyle\lim_{x\to3}$ during the computations (you can stop writing $\displaystyle\lim_{x\to3}$ when there is no longer $x$, i.e. after the substitution $x=3$), or writing things like $\displaystyle\lim_{x\to3}=\frac{\sqrt{3x}+3}{3}$ where $=$ sign is misplaced.
- Accidentally changing $\sqrt{3x}+3$ to $\sqrt{3x+3}$. Be careful with the length of the square root sign to avoid such mistakes.

Find all vertical asymptotes of the graph of the function

$\qquad\displaystyle f(x)=\frac{x^2+|x|-2}{x^2+x}$.

Justify your answer using limits.

**Answer:** $x=0$ is the only vertical asymptote.

Zeroes of the denominator are $x=0$ and $x=-1$. Show that $\displaystyle\lim_{x\to 0+} f(x)$ is infinite (as $x>0$, then $|x|=x$), and $\displaystyle\lim_{x\to -1} f(x)$ is finite (then $x<0$ and $|x|=-x$).

Zeros of the denominator: $x^2+x=0$ implies that $x=0$ or $x=-1$. There is no vertical asymptotes at other points as $f$ is continuous on its domain which is $\{x:x\ne0,-1\}$. At $x=0$:

$\qquad\displaystyle \lim_{x\to 0^+}\,\frac{x^2+|x|-2}{x^2+x}=\lim_{x\to 0^+}\,\frac{x^2+|x|-2}{x(x+1)} =-\infty \quad \left[\frac{-2}{0^+\cdot1}\right]$

Therefore, the graph of $f$ does have a vertical asymptote $x=0$.
$\qquad\displaystyle \lim_{x\to -1}\,\frac{x^2+|x|-2}{x^2+x}\underset{x<0}{=}\lim_{x\to -1}\,\frac{x^2-x-2}{x^2+x}=\lim_{x\to -1}\,\frac{(x+1)(x-2)}{x(x+1)}$

$\qquad\qquad\displaystyle =\lim_{x\to -1}\,\frac{x-2}x=\frac{-1-2}{-1}=3$.

Because $\displaystyle\lim_{x\to -1} f(x)$ is finite, the line $x=-1$ is not a vertical asymptote of the graph of $f(x)$.

- Incorrectly writing that $|x|=x$ as $x\to-1+$. It is true that $|x|=x$ as $x\to0+$ and that $|x|=-x$ as $x\to0-$. But one needs to be careful replacing $0$ with $-1$ in this argument. Namely, it is true that $|x|=-x$ as $x\to-1-$, but actually $|x|=-x$ even as $x\to-1+$ because when $x$ is close to $-1$ (no matter on which side), then $x<0$ (for example, think of $-0.9$ which is to the right from $-1$ and is negative). In turn, this mistake would lead to an infinite limit as $x\to-1+$ and incorrect conclusion that $x=-1$ is a vertical asymptote.
- Not a mistake, but unnecessary work. To establish that $x=0$ is a vertical asymptote, it is enough to establish that one of $\displaystyle\lim_{x\to 0+} f(x)$ or $\displaystyle\lim_{x\to 0-} f(x)$ is infinite. There is no need to consider both. (Note that to establish that $x=-1$ is NOT a vertical asymptote, both limits $\displaystyle\lim_{x\to -1-} f(x)$ and $\displaystyle\lim_{x\to -1+} f(x)$ or the two sided limit as in the above solution need to be considered.)