Limits and continuity

Why limits matter on this test

Calculus is about half of the Mathematics Test (ETS, GRE Mathematics Test fact sheet), and limits are its foundation. Limit questions are usually quick if you have the right tool in hand, and slow if you reach for L'Hôpital's rule by reflex. This lesson gives you a toolkit ordered from fastest to slowest.

Recall the definition: $\lim_{x\to a}f(x)=L$ means that for every $\varepsilon>0$ there is $\delta>0$ with $|f(x)-L|<\varepsilon$ whenever $0<|x-a|<\delta$. Note the $0<$: the value $f(a)$ itself is irrelevant to the limit.

Tool 1: the standard limits

Know these on sight (each follows from the power series in DLMF §4.6 or from the derivative at $0$):

Limit	Value
$\lim_{x\to0}\frac{\sin x}{x}$	$1$
$\lim_{x\to0}\frac{1-\cos x}{x^2}$	$\frac12$
$\lim_{x\to0}\frac{e^x-1}{x}$	$1$
$\lim_{x\to0}\frac{\ln(1+x)}{x}$	$1$
$\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x$	$e^a$
$\lim_{x\to0^+}x^p\ln x$ for $p>0$	$0$
$\lim_{n\to\infty}n^{1/n}$	$1$

Most textbook limits are one of these in disguise. For instance $\lim_{x\to 0}\frac{\sin 5x}{\tan 2x}=\lim \frac{\sin5x}{5x}\cdot\frac{2x}{\tan 2x}\cdot\frac52=\frac52$.

Tool 2: Taylor expansions (the fastest general method)

Replace each function by its Maclaurin polynomial to the order you need:

$$\sin x=x-\tfrac{x^3}{6}+O(x^5),\quad \cos x=1-\tfrac{x^2}{2}+\tfrac{x^4}{24}+O(x^6),\quad e^x=1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}+O(x^4),$$ $$\ln(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}+O(x^4),\quad \tan x=x+\tfrac{x^3}{3}+O(x^5),\quad (1+x)^a=1+ax+\tbinom{a}{2}x^2+O(x^3).$$

Worked example. $\displaystyle\lim_{x\to0}\frac{\tan x-\sin x}{x^3}$.

Numerator: $\left(x+\frac{x^3}{3}\right)-\left(x-\frac{x^3}{6}\right)+O(x^5)=\frac{x^3}{2}+O(x^5)$. The limit is $\frac12$. L'Hôpital would need three rounds of differentiation.

Worked example. $\displaystyle\lim_{x\to0}\frac{e^{x}-1-x-\frac{x^2}{2}}{x^3}=\frac16$, read straight from the series.

How far to expand: find the lowest power that survives in the denominator, and expand the numerator to that power.

Tool 3: L'Hôpital's rule, used correctly

If $f(x)\to0$ and $g(x)\to0$ (or both tend to $\pm\infty$) as $x\to a$, $f,g$ are differentiable near $a$ with $g'\neq0$ there, and $\lim f'/g'$ exists (finite or infinite), then $\lim f/g=\lim f'/g'$ (DLMF §1.4(iii), L'Hôpital's rule).

Three traps:

1. It is not indeterminate. $\lim_{x\to0}\frac{\cos x}{x+1}=1$; differentiating would give $\frac{-\sin 0}{1}=0$, which is wrong.
2. The derivative ratio has no limit. $\lim_{x\to\infty}\frac{x+\sin x}{x}=1$, but $\frac{1+\cos x}{1}$ oscillates. The rule then says nothing; divide by $x$ instead.
3. Circularity. Using L'Hôpital to prove $\frac{\sin x}{x}\to1$ assumes you know the derivative of $\sin$ at $0$, which is that very limit. On a multiple-choice test this is harmless; in a "which argument is valid" question it is the point.

Tool 4: indeterminate powers

For $1^\infty$, $0^0$ and $\infty^0$, write $f^g=e^{g\ln f}$ and find the limit of the exponent.

Worked example. $\displaystyle\lim_{x\to0}(\cos x)^{1/x^2}$. The exponent is $\frac{\ln\cos x}{x^2}$. Since $\cos x=1-\frac{x^2}{2}+\dots$ and $\ln(1+u)\approx u$, $\ln\cos x\approx-\frac{x^2}{2}$. The exponent tends to $-\frac12$; the limit is $e^{-1/2}$.

Shortcut for $1^\infty$: if $f\to1$ and $g\to\infty$, then $\lim f^g=e^{\lim g(f-1)}$ whenever the latter limit exists. Example: $\lim_{n\to\infty}\left(1+\frac{2}{n}+\frac{1}{n^2}\right)^{n}=e^{2}$.

Worked example. $\lim_{x\to0^+}x^x$: the exponent $x\ln x\to0$, so the limit is $e^0=1$.

Tool 5: limits of sums are integrals

If $f$ is integrable on $[0,1]$, then $\displaystyle\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}f\!\left(\frac kn\right)=\int_0^1 f(x)\,dx$ (Riemann integrals, DLMF §1.4(v)).

Worked example. $\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n}{n^2+k^2}=\lim\frac1n\sum_{k=1}^n\frac{1}{1+(k/n)^2}=\int_0^1\frac{dx}{1+x^2}=\frac\pi4.$

The recognition step is to factor out $\frac1n$ and see every remaining $k$ appear as $k/n$.

Tool 6: squeeze and growth rates

Squeeze: if $g\le f\le h$ near $a$ and $g,h\to L$, then $f\to L$. Example: $|x\sin(1/x)|\le|x|$, so $x\sin(1/x)\to0$ as $x\to0$.

Growth hierarchy as $n\to\infty$, for any $p>0$ and $a>1$:

$$\ln n\ \ll\ n^p\ \ll\ a^n\ \ll\ n!\ \ll\ n^n,$$

where $u\ll v$ means $u/v\to0$. So $\frac{n^{100}}{1.01^n}\to0$ and $\frac{n!}{n^n}\to0$. A useful refinement: $(n!)^{1/n}/n\to 1/e$ (from Stirling's formula, $n!\sim\sqrt{2\pi n}\,(n/e)^n$), so $(n!)^{1/n}$ grows like $n/e$.

Continuity

$f$ is continuous at $a$ when $\lim_{x\to a}f(x)=f(a)$ (DLMF §1.4(ii)). Discontinuities are removable (limit exists but differs from $f(a)$ or $f(a)$ undefined), jump (one-sided limits differ), or essential (a one-sided limit fails to exist, as for $\sin(1/x)$ at $0$).

Two theorems for continuous $f$ on a closed bounded interval $[a,b]$:

• Intermediate value theorem. $f$ takes every value between $f(a)$ and $f(b)$. Use it to show a root exists: $x^5-3x+1$ is $1$ at $0$ and $-1$ at $1$, so it has a root in $(0,1)$.
• Extreme value theorem. $f$ attains a maximum and a minimum.

Both fail on open intervals or for discontinuous functions, and the test likes to check that you know it.

Worked example (matching a piecewise function). Find $a,b$ such that

$$f(x)=\begin{cases}x^2, & x\le1\\ ax+b,& x>1\end{cases}$$

is differentiable at $1$. Continuity: $1=a+b$. Equal one-sided derivatives: $2=a$. So $a=2$, $b=-1$: the line is the tangent line at $x=1$. That is the general pattern: a piece joined differentiably to a curve is tangent to it.

Asymptotes, quickly

• Vertical: where the function blows up (a zero of the denominator that does not cancel).
• Horizontal: $\lim_{x\to\pm\infty}f(x)$, possibly different at $+\infty$ and $-\infty$ (e.g. $\arctan x$ has $y=\pm\frac\pi2$).
• Oblique: for a rational function whose numerator degree is one more than the denominator's, divide. $\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}$, so $y=x+1$.

Common traps

• Forgetting that $\lim_{x\to a}$ ignores $f(a)$.
• Taking $\sqrt{x^2}=x$ when $x\to-\infty$: $\lim_{x\to-\infty}\frac{\sqrt{x^2+1}}{x}=-1$, not $1$.
• Expanding to too low an order, so that everything cancels and you conclude $0$.
• Writing $1^\infty=1$. It is indeterminate: $(1+1/n)^n\to e$.
• Misreading a Riemann sum whose index runs to $2n$: then the interval is $[0,2]$.

Check yourself

• $\lim_{x\to0}\frac{1-\cos 3x}{x^2}=\ ?$ (Answer: $\frac92$.)
• $\lim_{x\to0}\frac{x-\sin x}{x^3}=\ ?$ (Answer: $\frac16$.)
• $\lim_{x\to\infty}\left(\frac{x+1}{x-1}\right)^x=\ ?$ (Answer: $e^2$.)
• $\lim_{n\to\infty}\frac1n\sum_{k=1}^n\sqrt{k/n}=\ ?$ (Answer: $\frac23$.)
• When does L'Hôpital's rule give no information even though the limit exists?
• Why does the extreme value theorem fail for $f(x)=x$ on $(0,1)$?