Can someone verify the following proofs?
Prove following functions have limit of $0$ as $\textbf{x} \rightarrow 0$
i) $f(x,y) = \frac{xy}{x+y+1}$
ii) $f(x,y) = \frac{\sin(x^2 + y^2)}{x^2 + y^2}$
iii) $f(x,y) = \frac{x^3}{x^2 + y^2}$
iv) $f(x,y) = \frac{x \sin^2 y}{x^+y^2}$
v) $f(x,y) = \frac{x^2 - y^2}{x - y}$ if $x \neq y$ and $0$ otherwise
Proofs:
i) assume that $|\textbf{x}| < \sqrt{\epsilon} \implies x < \sqrt{\epsilon} , y < \sqrt{\epsilon}$ now $|\frac{xy}{x+y+1}| \leq |xy| < \sqrt{\epsilon} . \sqrt{\epsilon} = \epsilon$ so $\delta = \sqrt{\epsilon}$ suffices
ii) since $\lim_{c \rightarrow 0} \frac{\sin c}{c} = 0$ there exists $\delta_1$ s.t $|c| < \delta \implies |\frac{\sin c}{c}| < \epsilon$. then we need $||\textbf{x}||^2 < \delta$ from which $||\textbf{x}|| < \sqrt{\delta_1}$ so our required $\delta = \sqrt{\delta_1} $
iii) since $|\frac{x^3}{x^2 + y^2}| \leq |x^3|$ choosing $||\textbf{x}|| < \epsilon^{1/3}$ implies that $|x|< \epsilon^{1/3}$ from which we will be done so $\delta = \epsilon^{1/3}$
iv) since $|\frac{x \sin^2 y}{x^2 + y^2}| \leq |x|$ and since $|x| < ||\textbf{x}||$ choosing $\delta = \epsilon$ sufficies
v) since $|\frac{x^2 - y^2}{x-y}| = |x + y| \leq |x| + |y|$ choosing $||\textbf{x}|| < \epsilon/2$ will give us $|x| < \epsilon$ and $|y| < \epsilon$ so $\delta = \epsilon/2$ suffices