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Question Number 145791 by mathmax by abdo last updated on 08/Jul/21

find lim_(x→0) ∫_0 ^x ((e^t +e^(−t) −2)/(1−cosx))dx

$$\mathrm{find}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{e}^{\mathrm{t}} +\mathrm{e}^{−\mathrm{t}} −\mathrm{2}}{\mathrm{1}−\mathrm{cosx}}\mathrm{dx} \\ $$

Commented by mathmax by abdo last updated on 09/Jul/21

lim_(x→0) ∫_0 ^x ((e^t +e^(−t) −2)/(1−cosx))dt

$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{e}^{\mathrm{t}} +\mathrm{e}^{−\mathrm{t}} −\mathrm{2}}{\mathrm{1}−\mathrm{cosx}}\mathrm{dt} \\ $$

Commented by mathmax by abdo last updated on 10/Jul/21

let use hospital rule =lim_(x→0) (1/(1−cosx))(e^x +e^(−x) −2) =lim_(x→0) ((e^x −e^(−x) )/(sinx)) =lim_(x→0) (((1+x+(x^2 /2))−(1−x +(x^2 /2)))/(sinx))=lim_(x→0) ((2x)/(sinx))=2

$$\mathrm{let}\:\mathrm{use}\:\mathrm{hospital}\:\mathrm{rule} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cosx}}\left(\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} −\mathrm{2}\right)\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{sinx}} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\left(\mathrm{1}−\mathrm{x}\:+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\mathrm{sinx}}=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{2x}}{\mathrm{sinx}}=\mathrm{2} \\ $$

Answered by Olaf_Thorendsen last updated on 09/Jul/21

f(x) = ∫_0 ^x ((e^t +e^(−t) −2)/(1−cosx)) dx f(x) = [((2sht−2t)/(1−cosx))]_0 ^x f(x) = ((2shx−2x)/(1−cosx)) f(x) ∼_0 ((2(x+(x^3 /(3!)))−2x)/(1−(1−(x^2 /(2!))))) ∼_0 (x/3) → 0

$${f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} \frac{{e}^{{t}} +{e}^{−{t}} −\mathrm{2}}{\mathrm{1}−\mathrm{cos}{x}}\:{dx} \\ $$$${f}\left({x}\right)\:=\:\left[\frac{\mathrm{2sh}{t}−\mathrm{2}{t}}{\mathrm{1}−\mathrm{cos}{x}}\right]_{\mathrm{0}} ^{{x}} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{2sh}{x}−\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}{x}} \\ $$$${f}\left({x}\right)\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{2}\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\right)−\mathrm{2}{x}}{\mathrm{1}−\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}\right)}\:\underset{\mathrm{0}} {\sim}\:\frac{{x}}{\mathrm{3}}\:\rightarrow\:\mathrm{0} \\ $$

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