Limits & Continuity in Two Variables

Top: $1 \cdot 2 + 3 = 5$. Bottom: $1 + 2 = 3 \ne 0$.

Not $\tfrac{0}{0}$: nothing to simplify, since the function is continuous at $(1, 2)$.

The limit is $\tfrac{5}{3}$.

$\dfrac{1 - 1}{1 - 1} = \dfrac{0}{0}$.

$\dfrac{(x - y)(x + y)}{x - y} = x + y$ wherever $x \ne y$ (where the function is defined).

$x + y \to 1 + 1 = 2$.

$\dfrac{0}{\sqrt{1} - 1} = \dfrac{0}{0}$.

Multiply top and bottom by $\sqrt{x^2 + y^2 + 1} + 1$. The bottom becomes $x^2 + y^2$, which cancels. What is left is $\sqrt{x^2 + y^2 + 1} + 1$.

$\sqrt{0 + 1} + 1 = 2$.

Top: $\sin 0 \cdot (1 - \cos 0) = 0$. Bottom: $0$. A $\tfrac{0}{0}$ form.

Split: $\dfrac{\sin(3x)}{x} \cdot \dfrac{1 - \cos y}{y^2}$. As $x \to 0$, $\dfrac{\sin(3x)}{x} = 3\,\dfrac{\sin(3x)}{3x} \to 3$. As $y \to 0$, $\dfrac{1 - \cos y}{y^2} \to \tfrac{1}{2}$.

Both limits exist, so the limit is $3 \cdot \tfrac{1}{2} = \tfrac{3}{2}$.

$y = 0$: $\dfrac{2x^2}{x^2} = 2 \to 2$. $\ x = 0$: $\dfrac{-y^2}{3y^2} = -\tfrac{1}{3} \to -\tfrac{1}{3}$.

Not needed: the axes already disagree.

Not needed: the axes already disagree.

Along the $x$-axis the limit is $2$; along the $y$-axis it is $-\tfrac{1}{3}$. Different, so no limit.

$y = 0$: $0$. $\ x = 0$: $0$. The axes agree, so try more.

$y = mx$: $\dfrac{4mx^2}{x^2 + 4m^2x^2} = \dfrac{4m}{1 + 4m^2}$, which depends on $m$. $m = 0$ gives $0$; $m = \tfrac{1}{2}$ gives $1$.

Not needed: two lines already disagree.

Along $y = 0$ the limit is $0$; along $y = \tfrac{x}{2}$ it is $1$. No limit.

$y = 0$: $0$. $\ x = 0$: $0$.

$y = mx$ ($m \ne 0$): $\dfrac{mx^4}{x^6 + m^2x^2} = \dfrac{mx^2}{x^4 + m^2} \to 0$. Every line gives $0$, which proves nothing.

Balance $x^6$ against $y^2$: take $y = x^3$. Then $\dfrac{x^6}{x^6 + x^6} = \tfrac{1}{2} \to \tfrac{1}{2}$.

Along $y = 0$ the limit is $0$; along $y = x^3$ it is $\tfrac{1}{2}$. No limit.

The lines through $(2, 0)$ parallel to the axes: $y = 0$ gives $0$; $x = 2$ gives $0$.

$y = m(x - 2)$: $\dfrac{m(x - 2)^2}{(1 + m^2)(x - 2)^2} = \dfrac{m}{1 + m^2}$. $m = 1$ gives $\tfrac{1}{2}$.

Not needed: two lines already disagree.

Along $y = 0$ the limit is $0$; along $y = x - 2$ it is $\tfrac{1}{2}$. No limit.

$\dfrac{3r^3\cos\theta\sin^2\theta}{r^2}$.

$= 3r\cos\theta\sin^2\theta$, so $L = 0$ and $p = 1$.

$|3\cos\theta\sin^2\theta| \le 3$.

$0 \le |f| \le 3r \to 0$, so the limit is $0$.

$\dfrac{r^2 + r^3\cos^2\theta\sin\theta}{r^2}$.

$= 1 + r\cos^2\theta\sin\theta$, so $L = 1$.

$|\cos^2\theta\sin\theta| \le 1$.

$|f - 1| \le r \to 0$, so the limit is $1$.

$\dfrac{r^4(\cos^4\theta + \cos^2\theta\sin^2\theta + \sin^4\theta)}{r^2}$.

$= r^2(\cos^4\theta + \cos^2\theta\sin^2\theta + \sin^4\theta)$, so $L = 0$.

$\cos^4\theta + \cos^2\theta\sin^2\theta + \sin^4\theta = 1 - \cos^2\theta\sin^2\theta \le 1$.

$0 \le f \le r^2 \to 0$, so the limit is $0$.

$\dfrac{r^2(\cos^2\theta - \cos\theta\sin\theta)}{r^2}$.

$= \cos^2\theta - \cos\theta\sin\theta$: no power of $r$ is left.

The $\theta$ part does not shrink: $\theta = 0$ gives $1$, $\theta = \tfrac{\pi}{2}$ gives $0$.

The value depends on the direction of approach, so there is nothing to squeeze: no limit.

$f(0, 0) = 0$.

Polar: $\dfrac{r^4\cos^2\theta\sin^2\theta}{r^2} = r^2\cos^2\theta\sin^2\theta$, and $0 \le f \le r^2 \to 0$. The limit is $0$.

Limit $0$ equals $f(0, 0) = 0$.

$f(0, 0) = 0$.

$y = 0$: $0$. $\ y = x$: $\dfrac{x^2}{3x^2} = \tfrac{1}{3}$. Two paths, two answers: the limit does not exist.

No limit, so nothing can equal $f(0, 0)$.

$f(0, 0) = c$.

Polar: $\dfrac{e^{r^2} - 1}{r^2}$. With $u = r^2 \to 0^+$, $\dfrac{e^u - 1}{u} \to 1$. The limit is $1$.

Continuous exactly when $c = 1$.