Limits, continuity and differentiability in depth

Limits

Definition. $\lim_{x \to a} f(x) = L$ means that $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a$, with $x \ne a$. Formally: for every $\varepsilon > 0$ there is a $\delta > 0$ such that $0 < \lvert x - a\rvert < \delta$ implies $\lvert f(x) - L\rvert < \varepsilon$.

One-sided limits. The limit exists exactly when the left limit $\lim_{x \to a^-} f(x)$ and the right limit $\lim_{x \to a^+} f(x)$ both exist and are equal. The value $f(a)$ itself plays no part.

Example 1. $f(x) = \dfrac{\lvert x\rvert}{x}$ has left limit $-1$ and right limit $+1$ at 0, so $\lim_{x\to 0} f(x)$ does not exist.

Standard limits to memorise

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

Growth order. As $x \to \infty$: $\ln x \ll x^p \ll a^x \ll x! \ll x^x$, for any $p > 0$ and $a > 1$.

Techniques for indeterminate forms

The forms $\frac00$, $\frac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$ and $\infty^0$ need work.

1. Factor and cancel. $\lim_{x\to3}\frac{x^2 - 9}{x - 3} = \lim_{x\to3}(x + 3) = 6 .$

2. Rationalise. $\lim_{x\to0}\frac{\sqrt{x + 4} - 2}{x} = \lim_{x\to0}\frac{(x + 4) - 4}{x(\sqrt{x + 4} + 2)} = \frac{1}{4}.$

3. Divide by the highest power for rational functions as $x \to \infty$. $\lim_{x\to\infty}\frac{3x^2 + 2}{5x^2 - x} = \frac35 .$ If the degrees differ, the limit is 0 (top degree smaller) or $\pm\infty$ (top degree larger).

4. L'Hôpital's rule. If the limit has the form $\frac00$ or $\frac{\infty}{\infty}$, then $\lim \frac{f}{g} = \lim \frac{f'}{g'}$, provided the second limit exists. The rule can be applied again if the form persists.

• $\lim_{x\to0}\dfrac{e^{2x} - 1}{\sin 3x} = \lim_{x\to0}\dfrac{2e^{2x}}{3\cos 3x} = \dfrac23$.
• $\lim_{x\to0}\dfrac{1 - \cos x}{x^2} = \lim\dfrac{\sin x}{2x} = \dfrac12$.

5. Rewrite $0 \cdot \infty$ as a quotient. $\lim_{x\to0^+} x\ln x = \lim\frac{\ln x}{1/x} = \lim\frac{1/x}{-1/x^2} = \lim(-x) = 0 .$

6. Exponential forms. Take logarithms: $y = f^g$ gives $\ln y = g \ln f$.

• $\lim_{x\to0^+} x^x = e^{\lim x\ln x} = e^0 = 1$.
• The $1^\infty$ shortcut: if $f \to 1$ and $g \to \infty$, then $\lim f^g = e^{\lim (f - 1)g}$.
• $\lim_{x\to\infty}\left(1 + \frac3x\right)^x = e^{3}$.
• $\lim_{x\to0}(\cos x)^{1/x^2} = e^{\lim (\cos x - 1)/x^2} = e^{-1/2}$.

7. Squeeze theorem. If $g \le f \le h$ near $a$, and $g$ and $h$ have the same limit $L$, then $f \to L$ too.

• $\lim_{x\to0} x\sin\frac1x = 0$, because $-\lvert x\rvert \le x\sin\frac1x \le \lvert x\rvert$.
• But $\lim_{x\to0}\sin\frac1x$ does not exist: the function oscillates between $-1$ and 1.

8. Combine fractions for $\infty - \infty$. $\lim_{x\to0}\left(\frac1x - \frac{1}{\sin x}\right) = \lim\frac{\sin x - x}{x\sin x} = 0 ,$ since the top is about $-x^3/6$ and the bottom about $x^2$.

Exam trap. L'Hôpital needs a genuine $\frac00$ or $\frac{\infty}{\infty}$ form. For $\lim_{x\to1}\frac{x^2 + 1}{x + 1}$, simply substitute: the answer is $\frac22 = 1$.

Continuity

Definition. $f$ is continuous at $a$ if three things hold:

1. $f(a)$ is defined;
2. $\lim_{x\to a} f(x)$ exists;
3. the limit equals $f(a)$.