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Question Number 209999 by som(math1967) last updated on 28/Jul/24

$$\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{{\mathit{e}}^{\mathit{x}}-1}{\sqrt{1-\mathit{c}\mathit{o}\mathit{s}\mathit{x}}}=?$$

Answered by RabieIsmail last updated on 28/Jul/24

$$\sqrt{2}$$

Commented by som(math1967) last updated on 28/Jul/24

$$islimitexist?$$$$$$

Commented by RabieIsmail last updated on 28/Jul/24

$$notexist$$

Commented by som(math1967) last updated on 28/Jul/24

$$thankyou$$

Answered by MM42 last updated on 28/Jul/24

$${lim}_{x\to 0}\frac{e^x-1}{\sqrt{1-cosx}}$$$$={lim}_{x\to 0}\frac{e^x-1}{\sqrt{2}\mid sin\frac{x}{2}\mid }$$$$=\{\begin{array}{l}{lim}_{x\to 0^{+}}\frac{e^x-1}{\sqrt{2}sin\frac{x}{2}}\stackrel{hop}{=}{lim}_{x\to 0^{+}}\frac{2e^x}{\sqrt{2}cos\frac{x}{2}}=\sqrt{2}\\ {lim}_{x\to 0^{-}}\frac{e^x-1}{-\sqrt{2}sin\frac{x}{2}}\stackrel{hop}{=}{lim}_{0^{-}}\frac{-2e^x}{\sqrt{2}cos\frac{x}{2}}=-\sqrt{2}\end{array}$$$$\Rightarrow lim:noexist✓$$$$$$

Commented by som(math1967) last updated on 28/Jul/24

$$thankyousir$$

Commented by MM42 last updated on 28/Jul/24

$$\underset{―}{\underbrace{⋚}}$$

Answered by klipto last updated on 29/Jul/24

$$\mathbf{sol}$$$$\mathbf{what}\mathbf{is}\sqrt{1-\mathbf{cosx}}?$$$$\mathbf{cosx}={\mathbf{cos}}^2\frac{\mathbf{x}}{2}-{\mathbf{sin}}^2\frac{\mathbf{x}}{2}$$$$\frac{\mathbf{x}}{2}=\mathit{\theta }$$$${\mathbf{cos}}^2\mathit{\theta }+{\mathbf{sin}}^2\mathit{\theta }=1$$$$1-\mathbf{cosx}={\mathbf{sin}}^2\mathit{\theta }+{\mathbf{cos}}^2\mathit{\theta }-(-{\mathbf{sin}}^2\mathit{\theta }+{\mathbf{cos}}^2\mathit{\theta })$$$$1-\mathbf{cosx}=2{\mathbf{sin}}^2\mathit{\theta }$$$$1-\mathbf{cosx}=2{\mathbf{sin}}^2\frac{\mathbf{x}}{2}$$$$\sqrt{1-\mathbf{cosx}}=\sqrt{2}\mid \mathbf{sin}\frac{\mathbf{x}}{2}\mid$$$$\therefore {\mathbf{lim}}_{\mathbf{x}\to 0}\frac{{\mathbf{e}}^{\mathbf{x}}-1}{\sqrt{1-\mathbf{cosx}}}$$$$\mathbf{sol}$$$${\mathbf{e}}^{\mathbf{x}}=1+\mathbf{x}+\frac{{\mathbf{x}}^2}{2!}+\frac{{\mathbf{x}}^3}{3!}+...$$$$\mathbf{sin}\frac{\mathbf{x}}{2}=\frac{\mathbf{x}}{2}-\frac{{\mathbf{x}}^3}{2\times 3!}+...$$$${\mathbf{lim}}_{\mathbf{x}\to 0}\frac{{\mathbf{e}}^{\mathbf{x}}-1}{\sqrt{1-\mathbf{cosx}}}=\frac{1+\mathbf{x}+\frac{{\mathbf{x}}^2}{2!}+\frac{{\mathbf{x}}^3}{3!}+...-1}{\sqrt{2}(\frac{\mathbf{x}}{2}-\frac{{\mathbf{x}}^3}{2\times 3!}+...)}=\frac{\mathbf{x}+\frac{{\mathbf{x}}^2}{2!}+\frac{{\mathbf{x}}^3}{3!}+...}{\sqrt{2}(\frac{\mathbf{x}}{2}-\frac{{\mathbf{x}}^3}{2\times 3!}+...)}=\frac{\mathbf{x}(1+\frac{\mathbf{x}}{2!}+\frac{{\mathbf{x}}^2}{3!}...)}{\mathbf{x}\sqrt{2}(\frac{1}{2}-\frac{{\mathbf{x}}^2}{2\times 3!}...)}$$$$=\frac{1+\frac{\mathbf{x}}{2!}+\frac{{\mathbf{x}}^2}{3!}...}{\sqrt{2}(\frac{1}{2}-\frac{{\mathbf{x}}^2}{2\times 3!}...)},{\mathbf{lim}}_{\mathbf{x}\to 0}\frac{{\mathbf{e}}^{\mathbf{x}}-1}{\sqrt{1-\mathbf{cosx}}}=\frac{1}{\frac{\sqrt{2}}{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}✓$$$$\mathbf{also}:\mathbf{RHL}\cancel{=}\mathbf{LHL}\therefore \mathbf{the}\mathbf{L}\mathrm{imit}\mathbf{DNE}$$$$\mathbf{klipto}-\mathbf{quanta}⨄$$

Commented by som(math1967) last updated on 29/Jul/24

$$thankyou$$

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