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Question Number 211759 by liuxinnan last updated on 20/Sep/24
lim_(x→0) ((e^(sin x) −e^x )/(sin x−x))=?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\mathrm{sin}\:{x}} −{e}^{{x}} }{\mathrm{sin}\:{x}−{x}}=? \\ $$
Commented by liuxinnan last updated on 20/Sep/24
Answered by TonyCWX08 last updated on 20/Sep/24
It′s 1
$${It}'{s}\:\mathrm{1} \\ $$
Answered by TonyCWX08 last updated on 20/Sep/24
Using L′Hopital rule Find the derivative for the numerator and denominator: (d/dx)(e^(sin(x)) −e^x )=cos(x)e^(sin(x)) −e^x (d/dx)(sin x − x )=cos (x)−1 The limit now transform to lim_(x→0) (((cos(x)e^(sin(x)) −e^x )/(cos (x)−1))) lim_(x→0) (((cos(x)e^(sin(x)) −e^x )/(cos (x)−1))) But it′s still not defined, so we apply the rule again. (d/dx)(cos(x)e^(sin(x)) −e^x )=−sin(x)e^(sin(x)) +cos^2 (x)e^(sin(x)) −e^x (d/dx)(cos(x)−1)=−sin(x) The limit now transform to lim_(x→0) (((−sin(x)e^(sin(x)) +cos^2 (x)e^(sin(x)) −e^x )/(−sin(x)))) =lim_(x→0) (((sin(x)e^(sin(x)) −cos^2 (x)e^(sin(x)) +e^x )/(sin(x)))) Still undefined, apply rule again (d/dx)(sin(x)e^(sin(x)) −cos^2 (x)e^(sin(x)) +e^x )=−e^(sin(x)) cos(x)sin^2 (x)−e^(sin(x)) sin(2x)−e^(sin(x)) cos(x)sin(x)−e^x (d/dx)(sin(x))=cos(x) The limit now transform to lim_(x→0) (((−e^(sin(x)) cos(x)sin^2 (x)−e^(sin(x)) sin(2x)−e^(sin(x)) cos(x)sin(x)−e^x )/(cos(x))) Finally, substitute x=0 The limit is 1
$${Using}\:{L}'{Hopital}\:{rule} \\ $$$${Find}\:{the}\:{derivative}\:{for}\:{the}\:{numerator}\:{and}\:{denominator}: \\ $$$$\frac{{d}}{{dx}}\left({e}^{\mathrm{sin}\left({x}\right)} −{e}^{{x}} \right)=\mathrm{cos}\left({x}\right){e}^{\mathrm{sin}\left({x}\right)} −{e}^{{x}} \\ $$$$\frac{{d}}{{dx}}\left(\mathrm{sin}\:{x}\:−\:{x}\:\right)=\mathrm{cos}\:\left({x}\right)−\mathrm{1} \\ $$$$ \\ $$$${The}\:{limit}\:{now}\:{transform}\:{to} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{cos}\left({x}\right){e}^{\mathrm{sin}\left({x}\right)} −{e}^{{x}} }{\mathrm{cos}\:\left({x}\right)−\mathrm{1}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{cos}\left({x}\right){e}^{\mathrm{sin}\left({x}\right)} −{e}^{{x}} }{\mathrm{cos}\:\left({x}\right)−\mathrm{1}}\right) \\ $$$$ \\ $$$${But}\:{it}'{s}\:{still}\:{not}\:{defined},\:{so}\:{we}\:{apply}\:{the}\:{rule}\:{again}. \\ $$$$\frac{{d}}{{dx}}\left({cos}\left({x}\right){e}^{{sin}\left({x}\right)} −{e}^{{x}} \right)=−{sin}\left({x}\right){e}^{{sin}\left({x}\right)} +{cos}^{\mathrm{2}} \left({x}\right){e}^{{sin}\left({x}\right)} −{e}^{{x}} \\ $$$$\frac{{d}}{{dx}}\left({cos}\left({x}\right)−\mathrm{1}\right)=−{sin}\left({x}\right) \\ $$$$ \\ $$$${The}\:{limit}\:{now}\:{transform}\:{to} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{−{sin}\left({x}\right){e}^{{sin}\left({x}\right)} +{cos}^{\mathrm{2}} \left({x}\right){e}^{{sin}\left({x}\right)} −{e}^{{x}} }{−{sin}\left({x}\right)}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{sin}\left({x}\right){e}^{{sin}\left({x}\right)} −{cos}^{\mathrm{2}} \left({x}\right){e}^{{sin}\left({x}\right)} +{e}^{{x}} }{{sin}\left({x}\right)}\right) \\ $$$$ \\ $$$${Still}\:{undefined},\:{apply}\:{rule}\:{again} \\ $$$$\frac{{d}}{{dx}}\left({sin}\left({x}\right){e}^{{sin}\left({x}\right)} −{cos}^{\mathrm{2}} \left({x}\right){e}^{{sin}\left({x}\right)} +{e}^{{x}} \right)=−{e}^{{sin}\left({x}\right)} {cos}\left({x}\right){sin}^{\mathrm{2}} \left({x}\right)−{e}^{{sin}\left({x}\right)} {sin}\left(\mathrm{2}{x}\right)−{e}^{{sin}\left({x}\right)} {cos}\left({x}\right){sin}\left({x}\right)−{e}^{{x}} \\ $$$$\frac{{d}}{{dx}}\left({sin}\left({x}\right)\right)={cos}\left({x}\right) \\ $$$$ \\ $$$${The}\:{limit}\:{now}\:{transform}\:{to} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{−{e}^{{sin}\left({x}\right)} {cos}\left({x}\right){sin}^{\mathrm{2}} \left({x}\right)−{e}^{{sin}\left({x}\right)} {sin}\left(\mathrm{2}{x}\right)−{e}^{{sin}\left({x}\right)} {cos}\left({x}\right){sin}\left({x}\right)−{e}^{{x}} }{{cos}\left({x}\right)}\right. \\ $$$${Finally},\:{substitute}\:{x}=\mathrm{0} \\ $$$${The}\:{limit}\:{is}\:\mathrm{1} \\ $$
Answered by BHOOPENDRA last updated on 20/Sep/24
As we know lim_(x→0) ((e^x −1)/(x ))=1 so from there lim_(x→0) e^x lim_(x→0) (((e^(sinx−x) −1)/(sinx−x))) ⇒ e^0 ×1 =1
$${As}\:{we}\:{know}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −\mathrm{1}}{{x}\:}=\mathrm{1} \\ $$$${so}\:{from}\:{there}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{e}^{{x}} \:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{e}^{{sinx}−{x}} −\mathrm{1}}{{sinx}−{x}}\right)\:\:\: \\ $$$$\Rightarrow\:{e}^{\mathrm{0}} \:×\mathrm{1} \\ $$$$=\mathrm{1} \\ $$
Answered by mehdee7396 last updated on 20/Sep/24
e^(sinx) =1+x+(x^2 /2)+O(x^4 ) e^x =1+x+(x^2 /2)+(x^3 /(3!))+O(x^4 ) ⇒lim_(x→0) ((e^(sinx) −e^x )/(sinx−x))=lim_(x→0) ((−(1/6)x^3 )/(((−1)/6)x^3 )) =1 ✓
$${e}^{{sinx}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{O}\left({x}^{\mathrm{4}} \right) \\ $$$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+{O}\left({x}^{\mathrm{4}} \right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{e}^{{sinx}} −{e}^{{x}} }{{sinx}−{x}}={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }{\frac{−\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }\:=\mathrm{1}\:\checkmark \\ $$
Commented by liuxinnan last updated on 20/Sep/24
Thanks everyone.It helps me a lot.
$${Thanks}\:{everyone}.{It}\:{helps}\:{me}\:{a}\:{lot}. \\ $$

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