Limits

A limit describes what $f(x)$ approaches as $x$ approaches a point, without caring what happens at the point itself. Continuity, derivatives and definite integrals are all defined through limits, and in computer science limits compare growth rates of running times. The tasks are to evaluate a limit (including an indeterminate form), decide whether a limit exists from one-sided behaviour, or pick a constant that makes a limit take a given value.

None within this subject. Algebra, the graphs of the elementary functions, and the derivatives of $x^n$, $\sin x$, $\cos x$, $e^x$ and $\ln x$ are assumed.

Let $f$ be defined on an open interval around $a$, except possibly at $a$. Then $\lim_{x \to a} f(x) = L$ means

The condition $0 < |x - a|$ excludes $x = a$: the value $f(a)$, or whether $f(a)$ is defined at all, has no effect on the limit. A limit, when it exists, is unique.

Variants use the same pattern. $\lim_{x \to \infty} f(x) = L$ means $|f(x) - L| < \varepsilon$ for all sufficiently large $x$. $\lim_{x \to a} f(x) = \infty$ means $f(x)$ exceeds any bound $M$ once $x$ is close enough to $a$; such a limit "does not exist" as a real number, but the notation records how it fails.

Limit versus value. $\lim_{x \to a} f(x)$ and $f(a)$ are different objects. They agree exactly when $f$ is continuous at $a$, which is the subject of Continuity and Differentiability.

The left limit $\lim_{x \to a^-} f(x)$ uses only $x < a$; the right limit $\lim_{x \to a^+} f(x)$ uses only $x > a$. The two-sided limit exists if and only if both one-sided limits exist and are equal. Piecewise definitions, $|x|$, $\lfloor x \rfloor$ and $\sqrt{x}$ at 0 call for one-sided limits.

If $\lim f = L$ and $\lim g = M$ (both finite, same approach $x \to a$), then the limits of $f \pm g$, $fg$ and $cf$ are $L \pm M$, $LM$ and $cL$, and $\lim f/g = L/M$ provided $M \ne 0$. If $h$ is continuous at $L$, then $\lim h(f(x)) = h(L)$; this is what lets you move a limit inside $\sqrt{\ }$, $\ln$ or $e^{(\cdot)}$.

Squeeze theorem. If $g(x) \le f(x) \le h(x)$ near $a$ and $\lim g = \lim h = L$, then $\lim f = L$. Standard use: $|x \sin(1/x)| \le |x|$, so $x \sin(1/x) \to 0$ as $x \to 0$, although $\sin(1/x)$ itself has no limit there.

As $x \to 0$:

Also $\lim_{x \to \infty}(1 + k/x)^x = e^k$ and $\lim_{x \to a} \dfrac{x^n - a^n}{x - a} = n a^{n-1}$.

When the limit laws give $0/0$, $\infty/\infty$, $0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$ or $\infty^0$, the answer depends on how fast each part moves, and more work is needed. The tools, roughly in order of preference:

For the three exponent forms, write $f^g = e^{g \ln f}$ and find the limit of $g \ln f$, which is a $0 \cdot \infty$ form.

Suppose $f$ and $g$ are differentiable near $a$ (except possibly at $a$), $g'(x) \ne 0$ near $a$, and $f/g$ is of the form $0/0$ or $\infty/\infty$ at $a$. If $\lim f'(x)/g'(x)$ exists (finite, $+\infty$ or $-\infty$), then $\lim f(x)/g(x)$ equals it. The same holds for one-sided limits and for $x \to \pm\infty$.

The rule differentiates numerator and denominator separately; it is not the quotient rule. It only works in one direction: if $\lim f'/g'$ does not exist, nothing follows about $\lim f/g$ (Example 9).

If $f(x) \to 1$ and $g(x) \to \infty$, then

because $\ln f = \ln(1 + (f - 1)) \approx f - 1$ when $f$ is near 1. For instance $(1 + 2/x)^x \to e^{2}$.

As $x\to0$, expand numerator and denominator far enough to expose their leading nonzero powers and coefficients. Equal powers give the ratio of coefficients; a higher numerator power gives 0. A lower numerator power requires checking one-sided signs: the result may be $+\infty$, $-\infty$, or no two-sided limit.

As $x \to \infty$, for any constants $p > 0$ and $a > 1$:

So $\ln x \ll x^p \ll a^x \ll x! \ll x^x$. A rational function behaves like the ratio of its leading terms: $\lim_{x \to \infty} (2x^2 + 3x)/(5x^2 - 1) = 2/5$. These facts are the limit form of the comparisons in Asymptotic Notation: if $f(n)/g(n) \to 0$ then $f = o(g)$; if the ratio tends to a positive constant then $f = \Theta(g)$.

$\lim_{x \to 0} \dfrac{\sin 3x}{5x} = \dfrac{3}{5} \lim_{x \to 0} \dfrac{\sin 3x}{3x} = \dfrac{3}{5}$.

$1 - \cos x = 2\sin^2(x/2)$, so $\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}\left(\dfrac{\sin(x/2)}{x/2}\right)^2 \to \dfrac{1}{2}$.

$\lim_{x \to \infty}(1 + 2/x)^x = e^{\lim x \cdot (2/x)} = e^2$. Similarly $\lim_{x \to 0}(1 + \sin x)^{1/x} = e^{\lim (\sin x)/x} = e$.

$\lim_{x \to 0} \dfrac{e^x - 1 - x}{x^2}$ is $0/0$. L'Hôpital gives $\dfrac{e^x - 1}{2x}$, still $0/0$; once more gives $\dfrac{e^x}{2} \to \dfrac{1}{2}$. With the series: the numerator is $x^2/2 + x^3/6 + \cdots$, so the ratio tends to $1/2$.

$\lim_{x \to 0^+} x \ln x = \lim \dfrac{\ln x}{1/x}$, an $\infty/\infty$ form; L'Hôpital gives $\dfrac{1/x}{-1/x^2} = -x \to 0$. Hence $\lim_{x \to 0^+} x^x = e^{\lim x \ln x} = e^0 = 1$.

$|x|/x$ equals $-1$ for $x < 0$ and $1$ for $x > 0$. The left limit at 0 is $-1$, the right limit is $1$, so $\lim_{x \to 0} |x|/x$ does not exist.

$\lim_{x \to 0} \dfrac{x - \sin x}{x^3}$: the numerator is $x - (x - x^3/6 + x^5/120 - \cdots) = x^3/6 - \cdots$, so the limit is $1/6$. L'Hôpital three times gives the same: $\dfrac{1 - \cos x}{3x^2} \to \dfrac{\sin x}{6x} \to \dfrac{\cos x}{6} \to \dfrac{1}{6}$.

$\lim_{x \to \infty} \dfrac{x + \sin x}{x} = \lim (1 + \sin x / x) = 1$, since $|\sin x/x| \le 1/x$. The ratio of derivatives, $1 + \cos x$, oscillates between 0 and 2 and has no limit, so the rule's hypothesis fails and it says nothing; the limit still exists.

For which $k$ does $\lim_{x \to 0} \dfrac{\sin 2x + kx}{x^3}$ exist (finite)? The numerator is $(2 + k)x - \dfrac{8x^3}{6} + \cdots$. The $x$ term must vanish, so $k = -2$, and then the limit is $-8/6 = -4/3$.