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in Limit, continuity and differentiability by (65 points)
edited by

If lim x tends to 0(1/x^2 - e^x/(e^x-1)^2 = I then the value of 1/l is

\(\lim\limits_{x\to 0}(\frac1{x^2}-\frac{e^x}{e^x-1})^2\)

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l = \(\lim\limits_{x\to 0}(\frac1{x^2}-\frac{e^x}{e^x-1})^2\)

 = \(\lim\limits_{x\to 0}(\frac1{x^2}-\frac{e^x-1-x^2e^x}{x^2(e^x-1)})^2\)

 = \((\lim\limits_{x\to 0}\frac{e^x-1-x^2e^x}{x^2(e^x-1)})^2\) (0/0 type)

 = \((\lim\limits_{x\to 0}\frac{e^x-x^2e^x-2xe^x}{x^2e^x+2xe^x-2x})^2\)(By applying D.L.H. Rule)

 = \((\frac10)^2\) = ∞

\(\therefore\) \(\frac1l=\frac1\infty\) = 0

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