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Question Number 145363 by imjagoll last updated on 04/Jul/21
 Without L′Hopital rule   lim_(x→π/4)  (((√2) cos x−1)/(cot x−1)) =?
$$\:\mathrm{Without}\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{rule} \\ $$$$\:\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{x}−\mathrm{1}}{\mathrm{cot}\:\mathrm{x}−\mathrm{1}}\:=? \\ $$
Answered by liberty last updated on 04/Jul/21
 lim_(x→(π/4))  (((√2)cos x−1)/(cot x−1))   = lim_(x→(π/4))  ((((√2)cos x−1)sin x)/(cos x−sin x)).((((√2)cos x+1))/( (√2)cos x+1)).((cos x+sin x)/(cos x+sin x))  = lim_(x→(π/4))  (((2cos^2 x−1))/(cos^2 x−sin^2 x)).((cos x+sin x)/( (√2)cos x+1)).sin x  = (((1/( (√2)))+(1/( (√2))))/( (√2) .(1/( (√2)))+1)).(1/( (√2))) = ((2/( (√2)))/2).(1/( (√2)))=(1/2)
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\mathrm{cos}\:{x}−\mathrm{1}}{\mathrm{cot}\:{x}−\mathrm{1}}\: \\ $$$$=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{cos}\:{x}−\mathrm{1}\right)\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}.\frac{\left(\sqrt{\mathrm{2}}\mathrm{cos}\:{x}+\mathrm{1}\right)}{\:\sqrt{\mathrm{2}}\mathrm{cos}\:{x}+\mathrm{1}}.\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\mathrm{2cos}\:^{\mathrm{2}} {x}−\mathrm{1}\right)}{\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x}}.\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{2}}\mathrm{cos}\:{x}+\mathrm{1}}.\mathrm{sin}\:{x} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\:\sqrt{\mathrm{2}}\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{1}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:=\:\frac{\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}}{\mathrm{2}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by puissant last updated on 04/Jul/21
=lim_(x→(π/4))   ((−(√2)sin(x))/(−(1/(sin^2 (x)))))  =  lim_(x→(π/4))  −(((√2)sin(x))/1)×(−sin^2 (x))  =lim_(x→(π/4))  ((√2)sin(x))×lim_(x→(π/4))  ((1/2)(1−cos(2x)))  =1×(1/2)=  (1/2)....
$$=\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\:\frac{−\sqrt{\mathrm{2}}\mathrm{sin}\left(\mathrm{x}\right)}{−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}}\:\:=\:\:\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:−\frac{\sqrt{\mathrm{2}}\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{1}}×\left(−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right) \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\left(\sqrt{\mathrm{2}}\mathrm{sin}\left(\mathrm{x}\right)\right)×\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)\right)\right) \\ $$$$=\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}=\:\:\frac{\mathrm{1}}{\mathrm{2}}…. \\ $$
Answered by Olaf_Thorendsen last updated on 04/Jul/21
lim_(x→(π/4))  (((√2)cosx−1)/(cotx−1))  lim_(u→0)  (((√2)cos((π/4)−u)−1)/((((√2)cos((π/4)−u))/( (√2)sin((π/4)−u)))−1))  lim_(u→0)  ((cosu+sinu−1)/(((cosu+sinu)/( cosu−sinu))−1))  lim_(u→0)  ((1−(u^2 /2)+u−1)/(((1−(u^2 /2)+u)/( 1−(u^2 /2)−u))−1))  lim_(u→0)  ((u−(u^2 /2))/((1−(u^2 /2)+u−1+u+(u^2 /2))/( 1−(u^2 /2)−u)))  lim_(u→0)  ((u−(u^2 /2))/((2u)/( 1−(u^2 /2)−u)))  lim_(u→0)  ((1−(u/2))/2)×(1−u−(u^2 /2)) = (1/2)
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\mathrm{cos}{x}−\mathrm{1}}{\mathrm{cot}{x}−\mathrm{1}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−{u}\right)−\mathrm{1}}{\frac{\sqrt{\mathrm{2}}\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−{u}\right)}{\:\sqrt{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−{u}\right)}−\mathrm{1}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}{u}+\mathrm{sin}{u}−\mathrm{1}}{\frac{\mathrm{cos}{u}+\mathrm{sin}{u}}{\:\mathrm{cos}{u}−\mathrm{sin}{u}}−\mathrm{1}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+{u}−\mathrm{1}}{\frac{\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+{u}}{\:\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−{u}}−\mathrm{1}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}}{\frac{\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+{u}−\mathrm{1}+{u}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}}}{\:\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−{u}}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}}{\frac{\mathrm{2}{u}}{\:\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−{u}}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\frac{{u}}{\mathrm{2}}}{\mathrm{2}}×\left(\mathrm{1}−{u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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