Mathematics - Limits Question with Solution | TestHub

Here, two straight lines, $(\tan \alpha )x+(\sin \alpha )y=\alpha$ and

$(\alpha \mathrm{cosec}\alpha )x+(\cos \alpha )y=1$, have their point of intersection as:

$$x=\frac{\alpha \cos \alpha -\sin \alpha }{\sin \alpha -\alpha }\text{ and }y=\frac{\alpha -x\tan \alpha }{\sin \alpha }$$

when $\alpha \to 0$, we obtain the point P.

$$\begin{array}{rl}& \text{ i.e., }\underset{\alpha \to 0}{\lim }x=\underset{\alpha \to 0}{\lim }\frac{\alpha \cos \alpha -\sin \alpha }{\sin \alpha -\alpha }\left(\frac{0}{0}\text{ form }\right)\\ & =\underset{\alpha \to 0}{\lim }\frac{-\alpha \sin \alpha +\cos \alpha -\cos \alpha }{\cos \alpha -1}\text{ (applying L-Hospital’s rule) }\\ & =\underset{\alpha \to 0}{\lim }\frac{-\alpha \sin \alpha }{-2{\sin }^2\alpha /2}=\underset{\alpha \to 0}{\lim }\frac{\alpha \left(2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\right)}{2{\sin }^2\frac{\alpha }{2}}\end{array}$$

$$\begin{array}{rl}& \underset{\alpha \to 0}{\lim }\frac{\alpha }{\tan \alpha /2}=\underset{\alpha \to 0}{\lim }\frac{2\frac{\alpha }{2}}{\tan \frac{\alpha }{2}}=2\\ & \text{ Again, }\underset{\alpha \to 0}{\lim }y=\underset{\alpha \to 0}{\lim }\frac{\alpha -x\tan \alpha }{\sin \alpha }=\underset{x\to 0}{\lim }\left(\frac{\alpha }{\sin \alpha }-\frac{x}{\cos \alpha }\right)\\ & \underset{\alpha \to 0}{\lim }\frac{\alpha }{\sin \alpha }-\underset{\alpha \to 0}{\lim }\frac{x}{\cos \alpha }=1-2=-1\left[\because \underset{\alpha \to 0}{\lim }x=2\right]\\ & \Rightarrow \underset{\alpha \to 0}{\lim }y=-1\end{array}$$

Hence, in the limiting position, $P(2,-1)\Rightarrow h-k=2-(-1)=3$.