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$\displaystyle \lim_{x \to 0} \frac{x(e^x - 1) + 2(\cos x -1)}{x(1 - \cos x)}$ is __________
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$\lim_{x \to 0} \frac{x(e^x - 1) + 2(\cos x -1)}{x(1 - \cos x)} = \lim_{x \to 0} \frac{x(e^x - 1)}{x(1 - \cos x)} + \frac{2(\cos x -1)}{x(1 - \cos x)}$

$=\lim_{x \to 0} \frac{(e^x - 1)}{(1 - \cos x)} - \frac{2}{x}$

Applying L'Hospital's rule,

$=\lim_{x \to 0} \frac{(e^x)}{(\sin x)} - \frac{0}{1}$

$=\lim_{x \to 0} \frac{(e^x)}{(\cos x)} = 1$
1
Sir but there is no 0/0 or inf./inf. Form in 2/x
0

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Use LH rule:

First Derivative$:\dfrac{ [x(e^{x}) + (e^{x}-1) - 2(\sin x)]}{[x \sin x + (1 - \cos x)]}$

Second Derivative$: \dfrac{[xe^{x} + e^{x} + e^{x} - 2 \cos x]}{[x \cos x + \sin x + \sin x]}$

Third Derivative$: \dfrac {[x e^{x} + e^{x} + e^{x} + e^{x} + 2 \sin x]}{[-x \sin x + \cos x + \cos x + \cos x]}$

Put $x = 0: \dfrac{[0+1+1+1+0]}{[0+1+1+1]} =\dfrac {3}{3} = 1.$
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we apply l hospital rule because putting limit in question it gives 0/0 form

so after 3 times applying L'hospital rule expression will be (3ex+xex+2sinx)/(-xsinx+3cosx)

now by putting limit value it gives 3/3=1 

Answer:

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