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Question Number 82887 by M±th+et£s last updated on 25/Feb/20

find without using l′hopital lim_(x→0) ((ln(1+cos(x))−ln(2))/x)

$${find}\:{without}\:{using}\:{l}'{hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\left(\mathrm{1}+{cos}\left({x}\right)\right)−{ln}\left(\mathrm{2}\right)}{{x}} \\ $$

Commented by msup trace by abdo last updated on 25/Feb/20

we have cosx ∼1−(x^2 /2) ⇒ 2+cosx ∼2−(x^2 /2) ⇒ ln(1+cosx)∼ln(2)+ln(1−(x^2 /4)) ∼ln(2)−(x^2 /4) ⇒((ln(1+cosx)−ln2)/x) ∼−(x/4)→0 (x→0) ⇒ lim_(x→0) ((ln(1+cosx)−ln2)/x)=0

$${we}\:{have}\:{cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{2}+{cosx}\:\sim\mathrm{2}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{cosx}\right)\sim{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\sim{ln}\left(\mathrm{2}\right)−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\frac{{ln}\left(\mathrm{1}+{cosx}\right)−{ln}\mathrm{2}}{{x}} \\ $$$$\sim−\frac{{x}}{\mathrm{4}}\rightarrow\mathrm{0}\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{ln}\left(\mathrm{1}+{cosx}\right)−{ln}\mathrm{2}}{{x}}=\mathrm{0} \\ $$

Answered by TANMAY PANACEA last updated on 25/Feb/20

lim_(x→0) ((ln(((1+cosx)/2)))/x) lim_(x→0) ((ln(1+((1+cosx)/2)−1))/((((1+cosx)/2)−1)))×((((1+cosx)/2)−1)/x) t=(((1+cosx)/2)−1) when x→0 t→0 lim_(t→0) ((ln(1+t))/t)×lim_(x→0) ((−2sin(x/2)×sin(x/2))/((x/2)×2)) 1×−1×0=0 t lim_(x→0)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ln}\left(\frac{\mathrm{1}+{cosx}}{\mathrm{2}}\right)}{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}+{cosx}}{\mathrm{2}}−\mathrm{1}\right)}{\left(\frac{\mathrm{1}+{cosx}}{\mathrm{2}}−\mathrm{1}\right)}×\frac{\frac{\mathrm{1}+{cosx}}{\mathrm{2}}−\mathrm{1}}{{x}} \\ $$$${t}=\left(\frac{\mathrm{1}+{cosx}}{\mathrm{2}}−\mathrm{1}\right)\:\:{when}\:{x}\rightarrow\mathrm{0}\:\:\:{t}\rightarrow\mathrm{0} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}×{sin}\frac{{x}}{\mathrm{2}}}{\frac{{x}}{\mathrm{2}}×\mathrm{2}} \\ $$$$\mathrm{1}×−\mathrm{1}×\mathrm{0}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${t} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}} \\ $$

Commented by M±th+et£s last updated on 25/Feb/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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