If $x=t-\sin t, y=1-\cos t$ and $\frac{d^2y}{dx^2}=-1$ at $t=K, K>0$, then $\lim_{t \to K} \frac{y}{x} =$
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Remember the formula for the second derivative of a parametric equation: $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$. A common mistake is forgetting to divide by $dx/dt$. Using trigonometric half-angle identities often simplifies the differentiation process.