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Question Number 209999 by som(math1967) last updated on 28/Jul/24
lim_(x→0) ((e^x −1)/( (√(1−cosx)))) =?
$$\:\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{{cosx}}}}\:=? \\ $$
Answered by RabieIsmail last updated on 28/Jul/24
(√2)
$$\sqrt{\mathrm{2}} \\ $$
Commented by som(math1967) last updated on 28/Jul/24
is limit exist ?
$$\:{is}\:{limit}\:{exist}\:? \\ $$$$ \\ $$
Commented by RabieIsmail last updated on 28/Jul/24
not exist
$${not}\:{exist} \\ $$
Commented by som(math1967) last updated on 28/Jul/24
thank you
$${thank}\:{you} \\ $$
Answered by MM42 last updated on 28/Jul/24
lim_(x→0) ((e^x −1)/( (√(1−cosx)))) = lim_(x→0) ((e^x −1)/( (√( 2))∣sin(x/2)∣)) = { ((lim_(x→0^+ ) ((e^x −1)/( (√2)sin(x/2)))=^(hop) lim_(x→0^+ ) ((2e^x )/( (√2)cos(x/2)))=(√2))),((lim_(x→0^− ) ((e^x −1)/( −(√2)sin(x/2)))=^(hop) lim_0^− ((−2e^x )/( (√2)cos(x/2))) =−(√2))) :} ⇒ lim : no exist ✓
$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{e}^{{x}} −\mathrm{1}}{\:\sqrt{\mathrm{1}−{cosx}}}\: \\ $$$$=\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{e}^{{x}} −\mathrm{1}}{\:\sqrt{\:\mathrm{2}}\mid{sin}\frac{{x}}{\mathrm{2}}\mid}\: \\ $$$$=\begin{cases}{{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\frac{{e}^{{x}} −\mathrm{1}}{\:\sqrt{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}}\overset{{hop}} {=}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\frac{\mathrm{2}{e}^{{x}} }{\:\sqrt{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}=\sqrt{\mathrm{2}}}\\{{lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\frac{{e}^{{x}} −\mathrm{1}}{\:−\sqrt{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}}\overset{{hop}} {=}\:{lim}_{\mathrm{0}^{−} } \:\frac{−\mathrm{2}{e}^{{x}} }{\:\sqrt{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}\:=−\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\Rightarrow\:{lim}\::\:{no}\:{exist}\:\checkmark \\ $$$$ \\ $$
Commented by som(math1967) last updated on 28/Jul/24
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by MM42 last updated on 28/Jul/24
⋛
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$
Answered by klipto last updated on 29/Jul/24
sol what is (√(1−cosx))? cosx=cos^2 (x/2)−sin^2 (x/2) (x/2)=𝛉 cos^2 𝛉+sin^2 𝛉=1 1−cosx=sin^2 𝛉+cos^2 𝛉−(−sin^2 𝛉+cos^2 𝛉) 1−cosx=2sin^2 𝛉 1−cosx=2sin^2 (x/2) (√(1−cosx))=(√2)∣sin(x/2)∣ ∴lim_(x→0) ((e^x −1)/( (√(1−cosx)))) sol e^x =1+x+(x^2 /(2!))+(x^3 /(3!))+... sin(x/2)=(x/2)−(x^3 /(2×3!))+... lim_(x→0) ((e^x −1)/( (√(1−cosx))))=((1+x+(x^2 /(2!))+(x^3 /(3!))+...−1)/( (√2)((x/2)−(x^3 /(2×3!))+...)))=((x+(x^2 /(2!))+(x^3 /(3!))+...)/( (√2)((x/2)−(x^3 /(2×3!))+...)))=((x(1+(x/(2!))+(x^2 /(3!))...))/( x(√2)((1/2)−(x^2 /(2×3!))...))) =((1+(x/(2!))+(x^2 /(3!))...)/( (√2)((1/2)−(x^2 /(2×3!))...))),lim_(x→0) ((e^x −1)/( (√(1−cosx))))=(1/((√2)/2))=(2/( (√2)))=(√2)✓ also:RHL=LHL ∴the Limit DNE klipto−quanta⊎
$$\boldsymbol{\mathrm{sol}} \\ $$$$\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{cosx}}}? \\ $$$$\boldsymbol{\mathrm{cosx}}=\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}} \\ $$$$\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}=\boldsymbol{\theta} \\ $$$$\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\theta}+\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\theta}=\mathrm{1} \\ $$$$\mathrm{1}−\boldsymbol{\mathrm{cosx}}=\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\theta}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\theta}−\left(−\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\theta}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\theta}\right) \\ $$$$\mathrm{1}−\boldsymbol{\mathrm{cosx}}=\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\theta} \\ $$$$\mathrm{1}−\boldsymbol{\mathrm{cosx}}=\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}−\boldsymbol{\mathrm{cosx}}}=\sqrt{\mathrm{2}}\mid\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\mid \\ $$$$\therefore\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} \frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} −\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{cosx}}}} \\ $$$$\boldsymbol{\mathrm{sol}} \\ $$$$\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} =\mathrm{1}+\boldsymbol{\mathrm{x}}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{2}×\mathrm{3}!}+… \\ $$$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} \frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} −\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{cosx}}}}=\frac{\mathrm{1}+\boldsymbol{\mathrm{x}}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{3}!}+…−\mathrm{1}}{\:\sqrt{\mathrm{2}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{2}×\mathrm{3}!}+…\right)}=\frac{\boldsymbol{\mathrm{x}}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{3}!}+…}{\:\sqrt{\mathrm{2}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{2}×\mathrm{3}!}+…\right)}=\frac{\boldsymbol{\mathrm{x}}\left(\mathrm{1}+\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}!}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{3}!}…\right)}{\:\boldsymbol{\mathrm{x}}\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}!}…\right)} \\ $$$$=\frac{\mathrm{1}+\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}!}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{3}!}…}{\:\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}!}…\right)},\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} \frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} −\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{cosx}}}}=\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}\checkmark \\ $$$$\boldsymbol{\mathrm{also}}:\boldsymbol{\mathrm{RHL}}\cancel{=}\boldsymbol{\mathrm{LHL}}\:\therefore\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{L}}\mathrm{imit}\:\boldsymbol{\mathrm{DNE}} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}}\biguplus \\ $$
Commented by som(math1967) last updated on 29/Jul/24
thank you
$${thank}\:{you} \\ $$

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