I used the path $y=x$ and got that the limit equals $12$. How do I manipulate the original expression to prove that the limit actually equals $12$ for all paths? Thank you!
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1Welcome to math.SE!! To write math expressions you have to use $\mathrm\LaTeX$ and, to render it, you have to ouse MathJax. – manooooh Feb 15 '19 at 21:31
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The limit is not $12$ but $1/12$. – manooooh Feb 15 '19 at 21:44
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@manooh
Computing the limit of $(1-\cos x)/x^2$ without l'Hopital's rule is easy: $$ \lim_{x\to 0} \frac {1 - \cos x} {x^2} = \lim_{x\to 0} \frac {1 - (1 - 1/2x^2 + \mathcal o(x^3))} {x^2} = \frac 1 2 + \lim_{x\to 0} \frac {\mathcal o (x^3)}{x^2} = \frac 1 2 $$
enedil
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Good approach! Please do not make an answer just to remark something to a user; instead, make a comment or post a new complete answer. – manooooh Feb 15 '19 at 22:33
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I would go for the original limit. We have $$\lim_{(x,y)\to(0,0)}\frac{\sin x(1-\cos y)}{6xy^2}=\frac16\lim_{(x,y)\to(0,0)}\underbrace{\frac{\sin x}x}_{\text{$1$ (remar-}\\\text{kable limit)}}\underbrace{\frac{1-\cos y}{y^2}}_{\{0/0\}\;(1)},$$ where in $(1)$ we can use this helpful answer (Finding the limit of $(1−\cos x)/x^2$), so $$\lim_{y\to0}\frac{1-\cos y}{y^2}=\lim_{y\to0}\frac{(1-\cos y)(1+\cos y)}{y^2(1+\cos y)}=\lim_{y\to0}\frac{\sin^2y}{y^2}\frac1{1+\cos y}=\frac12.\tag*{(*)}\label{(*)}$$ Hence, the original limit can be written as $$\lim_{(x,y)\to(0,0)}\frac{\sin x(1-\cos y)}{6xy^2}=\frac161\frac12=\frac1{12}.$$
$\ref{(*)}$ Thank you egreg to mention an easier solution!
manooooh
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1It is simpler if you multiply numerator and denominator by $1+\cos y$. :-) – egreg Feb 15 '19 at 22:15
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