If $\left(a+\sqrt{2}\,b\cos x\right)\!\left(a-\sqrt{2}\,b\cos y\right)=a^2-b^2$, where $a>b>0$, then $\dfrac{dx}{dy}$ at $\!\left(\dfrac{\pi}{4},\dfrac{\pi}{4}\right)$ is: [If answer expressed as $(a+b)/(a-b)$, find $(a+b)^2$ when $a=5,b=2$]
Let f, g and h be the real valued functions defined on \mathbb{R} as f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 1, & x = 0 \end{cases}, g(x) = \begin{cases} \frac{\sin(x+1)}{(x+1)}, & x \neq -1 \\ 1, & x = -1 \end{cases} and h(x) = 2[x] - f(x), where [x] is the greatest integer \leq x. Then the value of \lim_{x \to 1} g(h(x-1)) is