limit of multivariable function $f\left(x,y,z\right)=\left(x+y+z\right)\sin\left(\frac{1}{x}\right)\sin\left(\frac{1}{y}\right)$ this is a question from my calculus homework:
Check if the following function has a limit in $(0,0,0)$ and if it has, find the value:
$$f\left(x,y,z\right)=\left(x+y+z\right)\sin\left(\frac{1}{x}\right)\sin\left(\frac{1}{y}\right)$$
I tried to use different approaches in order to show that the limits are not equal so the function doesn't have a limit, but obviously I always got to $0$
for example:
$$x=y=z: \lim _{\left(z,z,z\right)\to \left(0,0,0\right)}\left(z+z+z\right)\sin\left(\frac{1}{z}\right)\sin\left(\frac{1}{z}\right)=\lim_{\left(z,z,z\right)\to \:\left(0,0,0\right)}3z\:\sin^2\left(\frac{1}{z}\right)$$
since $\lim _{z\to 0}3z=0$ and $\sin(\frac{1}{z})$ is a bounded function:
$$\lim_{\left(z,z,z\right)\to \:\left(0,0,0\right)}3z\:\sin^2\left(\frac{1}{z}\right)=0$$
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as well, I think converting coordinates won't help m
I have no more idea then...
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sorry about my weak grammar
so thankful for any help or hint
 A: Proposition
Let us suppose that we are given $X\subseteq\textbf{R}^{3}$, $f:X\rightarrow\textbf{R}$, $g:X\rightarrow\textbf{R}$ and an adherent point $x_{0}\in X$. Thus, if $f(x)\rightarrow 0$ when $x\rightarrow x_{0}$ and $|g(x)|\leq M$, we conclude that
\begin{align*}
\lim_{x\rightarrow x_{0}}f(x)g(x) = 0
\end{align*} 
Proof
According to the definition of limit, for every $\varepsilon/M > 0$, there is a $\delta > 0$ such that
\begin{align*}
0 < \|x - x_{0}\| < \delta \Longrightarrow |f(x) - 0| < \varepsilon/M \Longrightarrow |f(x)g(x)| < M\times\frac{\varepsilon}{M} = \varepsilon
\end{align*}
Hence we conclude that $f(x)g(x)\rightarrow 0$ as $x\rightarrow x_{0}$.
Solution
At your case, $f(x,y,z) = x + y + z$, $\displaystyle g(x,y,z) = \sin\left(\frac{1}{x}\right)\sin\left(\frac{1}{y}\right)$ and $M = 1$.
Hopefully this helps.
A: $$0<\left|\left(x+y+z\right)\sin\left(\frac{1}{x}\right)\sin\left(\frac{1}{y}\right)\right|<|x+y+z|$$
Taking limit as $(x,y,z)\rightarrow (0,0,0)$ you get the required limit.
