12. limits — MATHS STD 11 Science — Question

$ = \mathop {\lim }\limits_{\alpha \to \pi /4} \,\left\{ {\frac{{\sqrt 2 \left( {\sin \alpha .\frac{1}{{\sqrt 2 }} - \cos \alpha .\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\alpha - \frac{\pi }{4}} \right)}}} \right\}$

$ = \sqrt 2 \,\mathop {\lim }\limits_{\alpha \to \pi /4} \,\frac{{\sin \,\left( {\alpha - \frac{\pi }{4}} \right)}}{{\left( {\alpha - \frac{\pi }{4}} \right)}} = \sqrt 2 \times 1 = \sqrt 2 $.

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{\alpha \to \pi /4} \,\frac{{\sin \alpha - \cos \alpha }}{{\alpha - (\pi /4)}} = \mathop {\lim }\limits_{\alpha \to \pi /4} \,\frac{{\cos \alpha + \sin \alpha }}{1} = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = \sqrt 2 $.