Epsilon-delta proof of a multivariate limit

Question

I'd like to find $$\lim_{(x,y,z)\to(-2,1,-1)} f(x,y,z)$$ where $$f(x,y,z) = \frac{\sin(x+4y+2z)}{(x+4y+2z)}$$ I know that the limit should be 1, and I have a feeling that I should attempt to find the limit by using the fact that $$\lim_{u\to0} \frac{\sin(u)}{u} = 1$$ but I don't know how to use this fact in the case of multivariable limits. How should I proceed?

Answer 1

Method 1 : You can use the fact (a bit more accurate) that $$ \sin(u)=u+ o(u)$$ (which is true for all $u\in \mathbb{R}$, but which have an interest only near $0$) so, you will have $$\frac{\sin(x+4y+2z)}{x+4y+2z}= \frac{x+4y+2z+o(x+4y+2z)}{x+4y+2z}= 1 + o(1)\rightarrow 0$$ (when $(x,y,z)\rightarrow (-2,1,-1)$).

Method 2 :

Let $\varepsilon >0 $.

Since $\lim \frac{\sin(u)}{u}=1$, $$\exists \delta_1>0, \quad |u|\leq \delta_1 \Rightarrow \Big|\frac{\sin(u)}{u}-1\Big| <\varepsilon $$

Since $f: (x,y,z)\mapsto x+4x+2z$ is continuous and $f(-2,1,-1)=0$, $$\exists \delta_2>0, \quad || (x,y,z)-(-2,1,-1)||\leq \delta_2 \Rightarrow |f(x,y,z)-f(-2,1,-1)|=|f(x,y,z)| <\delta_1 $$

So, there exists $\delta_2>0$ for every $(x,y,z)$ such that $|| (x,y,z)-(-2,1,-1)||\leq \delta_2$, ($\Rightarrow$ $|f(x,y,z)| <\delta_1$ $\Rightarrow$) $$ \Big| \frac{\sin(x+4y+2z)}{x+4y+2z}-1 \Big| =\Big| \frac{\sin(f(x,y,z) )}{f(x,y,z)}-1 \Big|< \varepsilon .$$

Answer 2

Take $\varepsilon>0$. Since $\lim_{u\to 0}\frac{\sin u}u=1$, there is a $\delta>0$ such that $0<|u|<\delta\implies\left|\frac{\sin u}u\right|<\varepsilon$. Now, if $\|(x,y,z)-(-2,1,-1)\|<\frac{\delta}7$, then$$|x+2|,|y-1|,|z+1|<\frac{\delta}7.$$So\begin{align}|x+4y+2z|&=\bigl|(x+2)+4(y-1)+2(z+1)\bigr|\\&\leqslant|x+2|+4|y-1|+2|z+1|\\&<\frac{\delta}7+4\frac{\delta}7+2\frac{\delta}7\\&=\delta,\end{align}and therefore, by the choice of $\delta$,$$\left|\frac{\sin(x+4y+2z)}{x+4y+2z}\right|<\varepsilon.$$