Question-56453

Question Number 56453 by wasim last updated on 16/Mar/19

Commented by wasim last updated on 16/Mar/19

$$\mathrm{please}\:\:\mathrm{solve}\:\mathrm{question}\:\mathrm{7} \\ $$$$ \\ $$$$\mathrm{thanks} \\ $$

Commented by maxmathsup by imad last updated on 16/Mar/19

we have f(x)=((cos^2 x−sin^2 x−1)/( (√(x^2 +1))−1)) if x≠0 and f(0)=a f continue at 0 ⇔lim_(x→0) f(x)=f(0)=a but we have f(x)=((cos(2x)−1)/( (√(1+x^2 ))−1)) and cos(2x) ∼1−((4x^2 )/2) =1−2x^2 (x→0) ⇒cos(2x)−1∼−2x^2 (√(1+x^2 ))∼ 1+(1/2)x^2 ⇒(√(1+x^2 ))−1 ∼ (x^2 /2) ⇒f(x) ∼−4 (x→0) ⇒ lim_(x→0) f(x)=−4 =a .

$${we}\:{have}\:{f}\left({x}\right)=\frac{{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}−\mathrm{1}}\:{if}\:{x}\neq\mathrm{0}\:\:{and}\:{f}\left(\mathrm{0}\right)={a} \\ $$$${f}\:{continue}\:{at}\:\mathrm{0}\:\Leftrightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{f}\left({x}\right)={f}\left(\mathrm{0}\right)={a}\:\:\:{but}\:{we}\:{have} \\ $$$${f}\left({x}\right)=\frac{{cos}\left(\mathrm{2}{x}\right)−\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}}\:\:\:\:{and}\:{cos}\left(\mathrm{2}{x}\right)\:\sim\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{cos}\left(\mathrm{2}{x}\right)−\mathrm{1}\sim−\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\sim\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:\Rightarrow\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}\:\sim\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{f}\left({x}\right)\:\sim−\mathrm{4}\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=−\mathrm{4}\:={a}\:. \\ $$

Answered by ajfour last updated on 16/Mar/19

lim_(x→0) [f(x)=((1−cos 2x)/(1−(√(x^2 +1))))]=a =lim_(x→0) ((2sin^2 x)/(1−(1+(x^2 /2))))=a =− 4lim_(x→0) (((sin x)/x))^2 =a ⇒ a=−4×1=−4 .

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[{f}\left({x}\right)=\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right]={a} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} {x}}{\mathrm{1}−\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)}={a} \\ $$$$\:\:\:=−\:\mathrm{4}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} ={a} \\ $$$$\Rightarrow\:\:\:\:\:{a}=−\mathrm{4}×\mathrm{1}=−\mathrm{4}\:. \\ $$

Commented by wasim last updated on 17/Mar/19

$$\mathrm{thanks} \\ $$

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