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Question Number 120425 by bobhans last updated on 31/Oct/20

lim_(x→0) ((2^x −cos x)/(sin x)) ?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{{x}} −\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:? \\ $$

Answered by john santu last updated on 31/Oct/20

lim_(x→0) ((2^x −cos x)/(sin x)) = lim_(x→0) (((2^x −cos x)x)/(xsin x)) = lim_(x→0) (((2^x −cos x)/x)).(x/(sin x)) = lim_(x→0) ((2^x −cos x)/x) . lim_(x→0) (x/(sin x)) = lim_(x→0) ((2^x ln 2+sin x)/1) . 1 = ln 2.

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{{x}} −\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}^{{x}} −\mathrm{cos}\:{x}\right){x}}{{x}\mathrm{sin}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{2}^{{x}} −\mathrm{cos}\:{x}}{{x}}\right).\frac{{x}}{\mathrm{sin}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{{x}} −\mathrm{cos}\:{x}}{{x}}\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{sin}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{{x}} \:\mathrm{ln}\:\mathrm{2}+\mathrm{sin}\:{x}}{\mathrm{1}}\:.\:\mathrm{1}\:=\:\mathrm{ln}\:\mathrm{2}. \\ $$

Answered by Lordose last updated on 31/Oct/20

lim_(x→0) ((2^x −cosx)/(sinx)) = lim_(x→0) ((2^x ln2−sinx)/1) = ln2

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{\mathrm{x}} −\mathrm{cosx}}{\mathrm{sinx}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{\mathrm{x}} \mathrm{ln2}−\mathrm{sinx}}{\mathrm{1}}\:=\:\mathrm{ln2} \\ $$

Answered by Dwaipayan Shikari last updated on 31/Oct/20

lim_(x→0) ((2^x −cosx)/(sinx))=lim_(x→0) ((2^x −1)/x) (cosx→1 sinx→x) =lim_(x→0) (((xlog(2)+1−1))/x)=log(2) lim_(x→0) a^x =xlog(a)+1

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} −{cosx}}{{sinx}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} −\mathrm{1}}{{x}}\:\:\:\:\:\:\left({cosx}\rightarrow\mathrm{1}\:\:\:\:\:{sinx}\rightarrow{x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({xlog}\left(\mathrm{2}\right)+\mathrm{1}−\mathrm{1}\right)}{{x}}={log}\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{a}^{{x}} ={xlog}\left({a}\right)+\mathrm{1} \\ $$

Answered by mathmax by abdo last updated on 31/Oct/20

2^x =e^(xln2) ∼1+xln2 ,cosx ∼1−(x^2 /2) sinx∼x−(x^3 /6) ⇒((2^x −cosx)/(sinx))∼((1+xln2−1+(x^2 /2))/(x−(x^3 /6))) =((ln2+(x/2))/(1−(x^2 /6))) ⇒lim_(x→0) ((2^x −cosx)/(sinx)) =ln(2)

$$\mathrm{2}^{\mathrm{x}} =\mathrm{e}^{\mathrm{xln2}} \:\sim\mathrm{1}+\mathrm{xln2}\:\:\:\:,\mathrm{cosx}\:\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{sinx}\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\frac{\mathrm{2}^{\mathrm{x}} −\mathrm{cosx}}{\mathrm{sinx}}\sim\frac{\mathrm{1}+\mathrm{xln2}−\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}}{\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}} \\ $$$$=\frac{\mathrm{ln2}+\frac{\mathrm{x}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{2}^{\mathrm{x}} −\mathrm{cosx}}{\mathrm{sinx}}\:=\mathrm{ln}\left(\mathrm{2}\right) \\ $$

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