lim
∫
∂
∑
∞
π
Spell 1: The Direct Substitution Charm
Evaluate \(\lim_{(x,y) \to (1,2)} g(x,y)\), if the limit exists, where
\(g(x,y) = \frac{3x^2 - xy}{x^2 + y^2 + 3}\).
This spell works through direct substitution since the denominator never vanishes.
We substitute x = 1 and y = 2 into the magical formula:
\(g(1,2) = \frac{3(1)^2 - (1)(2)}{(1)^2 + (2)^2 + 3} = \frac{3 - 2}{1 + 4 + 3}\)
\(= \frac{1}{8}\)
The magical limit reveals itself to be \(\frac{1}{8}\).
Spell 2: The Cosine Enchantment
Evaluate \(\lim_{(x,y) \to (0,0)} \cos \left( \frac{x^3 + y^2}{x + y + 2} \right)\). If the limit exists.
First, we examine the magical argument inside the cosine function.
\(\lim_{(x,y) \to (0,0)} \frac{x^3 + y^2}{x + y + 2} = \frac{0 + 0}{0 + 0 + 2} = 0\)
Now we cast the cosine spell on this result:
\(\cos(0) = 1\)
The magical limit reveals itself to be 1.
Spell 3: The Rationalization Rune
Let \(f(x,y) = \frac{y^2 - xy}{\sqrt{x} - \sqrt{y}}\) for \((x,y) \ne (0,0)\).
Show that \(\lim_{(x,y) \to (0,0)} f(x,y) = 0\).
We begin by casting the rationalization spell on the denominator:
\(f(x,y) = \frac{(y^2 - xy)(\sqrt{x} + \sqrt{y})}{x - y}\)
Now we factor the numerator:
\(= \frac{y(y - x)(\sqrt{x} + \sqrt{y})}{x - y} = \frac{-y(x - y)(\sqrt{x} + \sqrt{y})}{x - y}\)
The (x-y) terms cancel (for x≠y):
\(= -y(\sqrt{x} + \sqrt{y})\)
Now we approach the magical origin point (0,0):
\(\lim_{(x,y) \to (0,0)} -y(\sqrt{x} + \sqrt{y}) = -0(\sqrt{0} + \sqrt{0}) = 0\)
The spell reveals the limit to be 0.
Spell 4: The Exponential Sine Charm
Evaluate \(\lim_{(x,y) \to (0,0)} \cos \left( \frac{e^x \sin y}{y} \right)\), if the limit exists.
First, we examine the magical fraction inside the cosine.
We recall the fundamental limit \(\lim_{y \to 0} \frac{\sin y}{y} = 1\) and \(\lim_{x \to 0} e^x = 1\).
Therefore, our fraction becomes:
\(\lim_{(x,y) \to (0,0)} \frac{e^x \sin y}{y} = \lim_{x \to 0} e^x \times \lim_{y \to 0} \frac{\sin y}{y} = 1 \times 1 = 1\)
Now we apply the cosine enchantment:
\(\cos(1)\) (approximately 0.5403)
The magical limit reveals itself to be \(\cos(1)\).
Spell 5: The Path-Dependent Curse
Let \(g(x,y) = \frac{x^2 y}{x^4 + y^2}\) for \((x,y) \ne (0,0)\) and \(f(0,0) = 0\).
Part (i): Along straight paths y = mx
We substitute y = mx:
\(g(x,mx) = \frac{x^2 (mx)}{x^4 + (mx)^2} = \frac{mx^3}{x^4 + m^2x^2} = \frac{mx}{x^2 + m^2}\)
Approaching the origin:
\(\lim_{x \to 0} \frac{mx}{x^2 + m^2} = \frac{0}{0 + m^2} = 0\)
Part (ii): Along parabolic paths y = kx²
We substitute y = kx²:
\(g(x,kx^2) = \frac{x^2 (kx^2)}{x^4 + (kx^2)^2} = \frac{kx^4}{x^4 + k^2x^4} = \frac{k}{1 + k^2}\)
Approaching the origin:
\(\lim_{x \to 0} \frac{k}{1 + k^2} = \frac{k}{1 + k^2}\)
This reveals the curse of path-dependence - the limit changes based on our approach!
Spell 6: The Continuity Blessing
Show that \(f(x,y) = \frac{x^2 - y^2}{y^2 + 1}\) is continuous at every \((x,y) \in \mathbb{R}^2\).
A function receives the continuity blessing if the limit exists and matches the function value.
The denominator \(y^2 + 1 \geq 1\) for all real y, so it never vanishes.
Both numerator and denominator are polynomial spells, which are continuous everywhere.
The ratio of two continuous spells remains continuous where the denominator is non-zero.
Therefore, f(x,y) is blessed with continuity throughout \(\mathbb{R}^2\).
Spell 7: The Sine-Exponential Fusion
Let \(g(x,y) = \frac{e^y \sin x}{x}\), for \(x \ne 0\) and \(g(0,0) = 1\). Show that g is continuous at (0,0).
To bestow continuity at (0,0), we must show \(\lim_{(x,y) \to (0,0)} g(x,y) = g(0,0) = 1\).
First, along the path y=0:
\(\lim_{x \to 0} \frac{e^0 \sin x}{x} = \lim_{x \to 0} \frac{\sin x}{x} = 1\)
For general approach to (0,0):
\(\lim_{(x,y) \to (0,0)} \frac{e^y \sin x}{x} = \lim_{y \to 0} e^y \times \lim_{x \to 0} \frac{\sin x}{x} = 1 \times 1 = 1\)
Since the limit matches g(0,0), the continuity blessing is granted at (0,0).