Without-L-Hopital-rule-lim-x-pi-4-2-cos-x-1-cot-x-1-

Question Number 145363 by imjagoll last updated on 04/Jul/21

Without L^{'} Hopital rule

\lim_{x \to π / 4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} = ?

Answered by liberty last updated on 04/Jul/21

\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}

= \lim_{x \to \frac{π}{4}} \frac{(\sqrt{2} \cos x - 1) \sin x}{\cos x - \sin x} . \frac{(\sqrt{2} \cos x + 1)}{\sqrt{2} \cos x + 1} . \frac{\cos x + \sin x}{\cos x + \sin x}

= \lim_{x \to \frac{π}{4}} \frac{(2 \cos^{2} x - 1)}{\cos^{2} x - \sin^{2} x} . \frac{\cos x + \sin x}{\sqrt{2} \cos x + 1} . \sin x

= \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\sqrt{2} . \frac{1}{\sqrt{2}} + 1} . \frac{1}{\sqrt{2}} = \frac{\frac{2}{\sqrt{2}}}{2} . \frac{1}{\sqrt{2}} = \frac{1}{2}

Answered by puissant last updated on 04/Jul/21

= \lim_{x \to \frac{π}{4}} \frac{- \sqrt{2} \sin (x)}{- \frac{1}{\sin^{2} (x)}} = \lim_{x \to \frac{π}{4}} - \frac{\sqrt{2} \sin (x)}{1} \times (- \sin^{2} (x))

= \lim_{x \to \frac{π}{4}} (\sqrt{2} \sin (x)) \times \lim_{x \to \frac{π}{4}} (\frac{1}{2} (1 - \cos (2 x)))

= 1 \times \frac{1}{2} = \frac{1}{2} \dots .

Answered by Olaf_Thorendsen last updated on 04/Jul/21

\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}

\lim_{u \to 0} \frac{\sqrt{2} \cos (\frac{π}{4} - u) - 1}{\frac{\sqrt{2} \cos (\frac{π}{4} - u)}{\sqrt{2} \sin (\frac{π}{4} - u)} - 1}

\lim_{u \to 0} \frac{\cos u + \sin u - 1}{\frac{\cos u + \sin u}{\cos u - \sin u} - 1}

\lim_{u \to 0} \frac{1 - \frac{u^{2}}{2} + u - 1}{\frac{1 - \frac{u^{2}}{2} + u}{1 - \frac{u^{2}}{2} - u} - 1}

\lim_{u \to 0} \frac{u - \frac{u^{2}}{2}}{\frac{1 - \frac{u^{2}}{2} + u - 1 + u + \frac{u^{2}}{2}}{1 - \frac{u^{2}}{2} - u}}

\lim_{u \to 0} \frac{u - \frac{u^{2}}{2}}{\frac{2 u}{1 - \frac{u^{2}}{2} - u}}

\lim_{u \to 0} \frac{1 - \frac{u}{2}}{2} \times (1 - u - \frac{u^{2}}{2}) = \frac{1}{2}