Computing $\lim_{(x,y)\to (0,0)}\frac{\sin(x+y)}{x+y}$ I'm trying to compute the following limits and the textbook that I'm looking at suggested the following method.
$$\lim_{(x,y)\to (0,0)}\frac{\sin(x+y)}{x+y}$$
$$\lim_{(x,y)\to (0,0)}\frac{\sin(xy)}{xy}$$
Each of these limits are equal to $\lim_{t\to 0}\frac{\sin(t)}{t}=1$.
However, I'm curious how to analytically prove that such change in variables yield the correct value of the original limit. That is, when are we justified in changing a vector $(x,y)$ to a single variable $t$? In the single variable case, I'm familiar with such manipulations, but I have a bit of uneasiness in seeing such manipulations for the first time in multivariable case. I'd appreciate it if anyone could explain to me why and when such transformations are justified in the multivariate case.
 A: Also, you can think of $f(x,y)=\frac{sin(x+y)}{x+y}$ as the composition $f=g\circ h$ of the two functions: $h(x,y)=x+y$, $g(t)=\frac{sin(t)}{t}$. Since the first one is continuous at the origin and the second one by what you know about one variable, you get
$$
\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}g(h(x,y))=g\big(\lim_{(x,y)\to(0,0)}h(x,y)\big)=g(0)=0.
$$
Of course, it's essential to have the $h$. But often one can create it. For instance:
$$
\lim_{(x,y)\to(0,0)}\frac{x\sin(x^2)}{x^2+y^2}=
\lim_{(x,y)\to(0,0)}\frac{x^3\sin(x^2)}{(x^2+y^2)x^2}=
\lim_{(x,y)\to(0,0)}x\cdot\frac{x^2}{{x^2+y^2}}\cdot\frac{\sin(x^2)}{x^2},
$$
and here the first factor goes to $0$, the second is bounded (in absolute value by $1$) and the third goes to $1$ by the argument we are discussing here. In the end the limit is $0$.
A: All of these can be resolved using Taylor's theorem for sine.
Note that $\sin(x) = x - x^3 \cdot g(x)$ where $g(x)$ is some other infinitely differentiable function. You can be more explicit with the full expansion, but  that is not necessary.
Then $\sin(x+y)/(x+y) = 1 - (x+y)^2 \cdot g(x+y)$. Then as $(x,y) \to (0,0)$ we have $x+y \to 0$ and so the limit becomes $1- 0 \cdot g(0) = 1$.
An identical analysis can be performed for the second limit as well.
