Question-175160

Question Number 175160 by thean last updated on 21/Aug/22

Commented by Alisajadrajaee last updated on 21/Aug/22

lim_(x→(π/4)) ((((√2)cosx-1)/(tanx-1)))=(0/0) lim_(x→(π/4)) ((((√2)cosx-1)^′ )/((tanx-1)^′ ))=(((√2)(-sinx))/(sec^2 x)) lim_(x→(π/4)) -(((√2)(sin(π/4)))/(sec^2 (π/4)))=-(((√2)(((√2)/2)))/(((√2))^2 ))= -(1/2)

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{2}}{cosx}-\mathrm{1}}{{tanx}-\mathrm{1}}\right)=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\left(\sqrt{\mathrm{2}}{cosx}-\mathrm{1}\right)^{'} }{\left({tanx}-\mathrm{1}\right)^{'} }=\frac{\sqrt{\mathrm{2}}\left(-{sinx}\right)}{{sec}^{\mathrm{2}} {x}} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}-\frac{\sqrt{\mathrm{2}}\left({sin}\frac{\pi}{\mathrm{4}}\right)}{{sec}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}}=-\frac{\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)}{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\:-\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Answered by cortano1 last updated on 21/Aug/22

L=lim_(x→(π/4)) ((((√2) cos x−1)/(tan x−1))) L= lim_(x→(π/4)) ((((√2) (cos x−cos (π/4)))/((((sin (x−(π/4)))/(cos (π/4) cos x)))))) L= lim_(x→(π/4)) (((√2) .(1/( (√2))).cos x(−2sin (((x+(π/4))/2))sin (((x−(π/4))/2))))/(sin (x−(π/4)))) L= −((√2)/2).((√2)/2) .lim_(x→(π/4)) ((2sin (1/2)(x−(π/4)))/(sin (x−(π/4)))) L=−(1/4).2=((−1)/2)

$$\:\:{L}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\left(\frac{\sqrt{\mathrm{2}}\:\mathrm{cos}\:{x}−\mathrm{1}}{\mathrm{tan}\:{x}−\mathrm{1}}\right) \\ $$$$\:{L}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\left(\frac{\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:{x}−\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\right)}{\left(\frac{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:\mathrm{cos}\:{x}}\right)}\right) \\ $$$${L}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\mathrm{cos}\:{x}\left(−\mathrm{2sin}\:\left(\frac{{x}+\frac{\pi}{\mathrm{4}}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{x}−\frac{\pi}{\mathrm{4}}}{\mathrm{2}}\right)\right)}{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\:{L}=\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:.\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}\: \\ $$$$\:{L}=−\frac{\mathrm{1}}{\mathrm{4}}.\mathrm{2}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$

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