Limits

Get­ting close with­out arriv­ing#

What value does $\displaystyle \tfrac{x^2 - 9}{x - 3}$ approach as $x$ approaches $3$? At $x = 3$ itself the for­mula breaks down, since you would be divid­ing by zero. Yet if you try val­ues just a lit­tle either side, the answers set­tle qui­etly towards $6$. That value they set­tle on is called a limit, and it is the idea on which all of cal­cu­lus rests. We will look at left and right lim­its, learn the rules for com­bin­ing lim­its, and work out lim­its of poly­no­mial, ratio­nal and trigono­met­ric func­tions, includ­ing two results you will use again and again, $\displaystyle \lim_{x \to a}\tfrac{x^n - a^n}{x - a} = na^{n-1}$ and $\displaystyle \lim_{x \to 0}\tfrac{\sin x}{x} = 1$.

What a limit means#

$\displaystyle \lim_{x \to a} f(x) = L$ means you can make $f(x)$ as close to $L$ as you like by tak­ing $x$ close enough to $a$ (but never equal to it). Notice what this does not say: the value $f(a)$ itself plays no part at all.

One-sided lim­its. $\displaystyle \lim_{x \to a^-}$ looks only at $x < a$, while $\displaystyle \lim_{x \to a^+}$ looks only at $x > a$. The limit exists exactly when both of these exist and agree.

Alge­bra of lim­its#

Lim­its behave politely with arith­metic. If $\displaystyle \lim f = l$ and $\displaystyle \lim g = m$ at the same point, then $\displaystyle \lim(f \pm g) = l \pm m$; $\displaystyle \lim(fg) = lm$; $\displaystyle \lim \tfrac{f}{g} = \tfrac{l}{m}$ if $m \ne 0$; $\displaystyle \lim cf = cl$.

In prac­tice this means: for a poly­no­mial, $\displaystyle \lim_{x \to a} p(x) = p(a)$. For a ratio­nal func­tion whose denom­i­na­tor is not zero at $a$, just sub­sti­tute. If sub­sti­tut­ing gives $\displaystyle \tfrac{0}{0}$, do not panic; there is a com­mon fac­tor hid­ing, so can­cel it first.

A stan­dard limit#

$$\displaystyle \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$$

For pos­i­tive inte­gers $n$ you can prove this by fac­toris­ing $x^n - a^n$. It stays true for every ratio­nal $n$ pro­vided $a > 0$.

Trigono­met­ric lim­its#

For $\displaystyle 0 < x < \tfrac{\pi}{2}$, draw the unit cir­cle and com­pare three areas: a small tri­an­gle, a sec­tor, and a larger tri­an­gle. Their areas give $\displaystyle \cos x < \tfrac{\sin x}{x} < 1$. The mid­dle quan­tity is squeezed between the other two, so

$$\displaystyle \lim_{x \to 0}\frac{\sin x}{x} = 1, \qquad \lim_{x \to 0}\frac{1 - \cos x}{x} = 0.$$

Try these your­self#

1. Eval­u­ate: $\displaystyle \lim_{x \to -1}(2x^2 - 3x + 4)$; $\displaystyle \lim_{x \to 4} \frac{x^2 - 16}{x - 4}$; $\displaystyle \lim_{x \to 2} \frac{x^2 - 5x + 6}{x^2 - 4}$.
2. Eval­u­ate: $\displaystyle \lim_{x \to 1} \frac{x^7 - 1}{x - 1}$; $\displaystyle \lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}$; $\displaystyle \lim_{x \to 0} \frac{(1 + x)^6 - 1}{x}$.
3. Eval­u­ate: $\displaystyle \lim_{x \to 0} \frac{\sin 4x}{x}$; $\displaystyle \lim_{x \to 0} \frac{\tan 3x}{\sin 2x}$; $\displaystyle \lim_{x \to 0} \frac{1 - \cos 4x}{x^2}$.
4. Eval­u­ate: $\displaystyle \lim_{x \to 0} \frac{\sqrt{4 + x} - 2}{x}$; $\displaystyle \lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)}$.
5. Find the left and right lim­its at $0$ of $f(x) = \begin{cases} 2x + 3, & x \le 0 \\ 3(x + 1), & x > 0 \end{cases}$. Does the limit exist?
6. For what $k$ does $\displaystyle \lim_{x \to 2} f(x)$ exist if $f(x) = kx - 1$ for $x < 2$ and $x^2 + 1$ for $x \ge 2$?
7. Does $\displaystyle \lim_{x \to 0} [x]$ exist? What about $\displaystyle \lim_{x \to 0.5} [x]$?
8. Eval­u­ate $\displaystyle \lim_{x \to 0} \frac{x\cos x + \sin x}{x^2 + \tan x}$.
9. Eval­u­ate $\displaystyle \lim_{x \to a} \frac{\sin x - \sin a}{x - a}$.
10. Eval­u­ate $\displaystyle \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}$ for non-zero $a, b$ with $a + b \ne 0$.

Answers to check against#