Mathematics - Limits Question with Solution | TestHub

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

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

$$x=\frac{\alpha \cos \alpha -\sin \alpha }{\sin \alpha -\alpha }andy=\frac{\alpha -x\tan \alpha }{\sin \alpha }$$

∴ when $\alpha \to o$, we obtain the pointP.

$$\begin{array}{rlr}& i.e.,\underset{\alpha \to 0}{\lim }x=\underset{\alpha \to 0}{\lim }\frac{\alpha \cos \alpha -\sin \alpha }{\sin \alpha -\alpha }\left(\frac{0}{0}form\right)& \end{array}$$

$$=\underset{\alpha \to 0}{\lim }\frac{-\alpha \sin \alpha +\cos \alpha -\cos \alpha }{\cos \alpha -1}$$

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

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