Here are two examples (Examples 9.1 and 9.2) taken from Gelbaum and Olmsted's Counterexamples in Analysis that illustrate how badly "path checking" can fail:
1)
$$
f(x,y)=\cases{ {x^2y\over x^4+y^2 },& if $(x,y)\ne(0,0)$\cr 0,&$(x,y)=(0,0)$}
$$
Here the limit of $f$ as $(x,y)$ makes any straight line approach to the origin is $0$. Yet $f$ does not have a limit at $(0,0)$, as there are points arbitrarily near the origin at which $f$ takes the value $1/2$ (namely $(a,a^2)$.
2)
$$
f(x,y)=\cases{ {e^{-1/x^2}y\over e^{-2/x^2}+y^2 },& if $x\ne 0$\cr 0,&$x=0$}
$$
Here the limit of $f$ as $(x,y)$ makes any approach to the origin along a curve of the form $y=cx^{m/n}$, where $c\ne0$ and $m,n$ are relatively prime positive integers, is $0$. Yet $f$ does not have a limit at $(0,0)$, as there are points arbitrarily near the origin at which $f$ takes the value $1/2$ (namely $(a,e^{-1/a^2})$.
Best Answer
Consider the curves $\gamma_{1} = (t,t)$ and $\gamma_{2}(t) = (t,t^{2})$. Thus we have \begin{align*}f(\gamma_{1}(t)) = \frac{t^{2}\sin^{2}(t)}{t^{4} + t^{4}} = \frac{\sin^{2}(t)}{2t^{2}} \Longrightarrow \lim_{t\rightarrow 0}f(\gamma_{1}(t)) = \lim_{t\rightarrow 0}\left[\frac{1}{2}\left(\frac{\sin(t)}{t}\right)^{2}\right] = \frac{1}{2} \end{align*}
On the other hand, we have \begin{align*} f(\gamma_{2}(t)) = \frac{t^{8}\sin^{2}(t)}{t^{4} + t^{8}} = \frac{t^{4}\sin^{2}(t)}{1 + t^{4}} \Longrightarrow \lim_{t\rightarrow 0}f(\gamma_{2}(t)) = \lim_{t\rightarrow 0}\frac{t^{4}\sin^{2}(t)}{1+t^{4}} = \frac{0\times 0}{1+0} = 0 \end{align*}
If the given limit existed, we should have $\displaystyle\lim_{t\rightarrow0}f(\gamma_{1}(t)) = \lim_{t\rightarrow 0}f(\gamma_{2}(t))$. Hence $\displaystyle\lim_{\textbf{x}\rightarrow\textbf{0}} f(\textbf{x})$ does not exist. Hope this helps.