Using definition,How can I find that limit of below function of $\mathbb{R}^2$:
$1.\displaystyle \lim_{(x,y) \to (0,0)} \frac{\sin(x+y)}{(x+y)} $
$2. \displaystyle \lim_{(x,y) \to (0,0)} x \sin \bigg(\frac{1}{y}\bigg)+y \sin\bigg(\frac{1}{x}\bigg)$
Please help me, thanks in advance.
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For $(2)$, observe that $$|x| \le \sqrt{x^2+y^2}$$ and $$|y| \le \sqrt{x^2+y^2}$$. So given $\epsilon \gt 0$, whenever there exists a $\delta \gt 0$ such that $ \sqrt{x^2+y^2}\lt \delta$, we have $$\displaystyle | x \sin \bigg(\frac{1}{y}\bigg)+y \sin\bigg(\frac{1}{x}\bigg)-0|=\displaystyle | x \sin \bigg(\frac{1}{y}\bigg)+y \sin\bigg(\frac{1}{x}\bigg)| \le |x|+|y| \lt 2\delta$$
By choosing $\delta =\dfrac{\epsilon}{2}$, we will have the limit $0$.
For $(1)$, we will use the fact that $$\lim_{u \to 0} \frac{\sin u}{u}=1$$. From here, given $\epsilon \gt 0$, there exists a $\delta \gt 0$ such that we have $$0<|u|<2\delta \Rightarrow \left|\frac{\sin u}{u}-1\right|<\epsilon. \qquad (*)$$. Now , given $\epsilon \gt 0$, choose $\delta$ such that $(*)$holds.
Now let $$0<\sqrt{x^2+y^2}<\delta$$. Then $$|x+y|\leq|x|+|y|\leq\sqrt{|x|^2+|y|^2}+\sqrt{|x|^2+|y|^2}=2\sqrt{x^2+y^2}<2\delta.$$
by $(*) (for x+y\ne 0)$ and by the definition of $f $, we have $|f(x,y)−1|<ϵ$ as required.
Point 2 is trivial: just remark that the two terms are multiplication of an infinitesimal and a bounded quantity, so the limit is zero. More formally, pick $\epsilon>0$ and write $$ \left|x \sin \frac{1}{y} + y \sin \frac{1}{x} \right| \leq |x| + |y| < \epsilon $$ provided that $|x|<\frac{\epsilon}{2}$, $|y|<\frac{\epsilon}{2}$.
Point 1 is tricky, since it is just a restatement of a popular limit: $\lim_{z \to 0} \frac{\sin z}{z}=1$, with $z=x+y \to 0$. So the answer is: the limit is $1$, and the "elementary" proof can be read on most calculus books.