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problem on limits...

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problem on limits... Empty problem on limits...

Post  mohit Mon Jul 18, 2011 1:39 pm

Find the value of. problem on limits... Gif.latex?\mathbf{\lim_{x\rightarrow%200}\frac{e^{\tan%20x}-e^{\sin%20x}}{\left(e^{2x}-1\right)
(edited by moderator.)

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problem on limits... Empty Re: problem on limits...

Post  Admin Mon Jul 18, 2011 8:04 pm

problem on limits... Gif.latex?\mathbf{\lim_{x\rightarrow%200}\frac{e^{\tan%20x}-e^{\sin%20x}}{(e^{2x}-1).(1-\cos%204x)}}$\\\\\\\%20\textbf{Now%20We%20now%20that%20$\mathbf{\lim_{x\rightarrow%200}\frac{\left(e^{2x}-1\right)}{2x}=1}$}\\\\\\%20$\mathbf{\lim_{x\rightarrow%200}\frac{\left(1-\cos%204x\right)}{4x^2}=\lim_{x\rightarrow%200}\frac{2\sin^2%202x}{(2x)^2}=2}$\\\\\\%20\textbf{Now%20Here%20$\mathbf{\lim_{x\rightarrow%200}\frac{e^{\tan%20x}-e^{\sin%20x}}{(e^{2x}-1).(1-\cos%204x)}=\frac{e^{\tan%20x}-e^{\sin%20x}}{\lim_{x\rightarrow%200}\left(\frac{e^{2x}-1}{2x}\right).\lim_{x\rightarrow%200}\left(\frac{1-\cos%204x}{4x^2}\right).(2x).(4x^2)}}$}\\\\\\%20$\mathbf{\lim_{x\rightarrow%200}\frac{e^{\tan%20x}-e^{\sin%20x}}{8x^3}=\lim_{x\rightarrow%200}\frac{\left(\frac{e^{\tan%20x}-1}{\tan%20x}\right).\tan%20x-\left(\frac{e^{\sin%20x}-1}{\sin%20x}\right).\sin%20x}{8x^3}}$\\\\\\%20$\mathbf{\lim_{x\rightarrow%200}%20\frac{\tan%20x%20-\sin%20x}{8x^3}=\lim_{x\rightarrow%200}\frac{\sin%20x.(1-\cos%20x)}{\cos%20x.8x^3}=\lim_{x\rightarrow%200}\left(\frac{1-\cos%20x}{x^2.}\right).\frac{1}{8}
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problem on limits... Empty thks...............

Post  mohit Tue Jul 19, 2011 1:23 pm

thank you sir for giving solution.......

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