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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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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 

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Form :- (→0/→0) => indeterminate form.

Method 1 :- Use taylor series formula for e$^{x}$ and cos(x).

e$^{x}$ = 1 + x + x$^{2}$/2! + x$^{3}$/3! + x$^{4}$/4! + ….
=> e$^{x}$ – 1 = x + x$^{2}$/2! + x$^{3}$/3! + x$^{4}$/4! + …. = x( 1 + x/2! + x$^{2}$/3! + x$^{3}$/4! + ….)
=> x(e$^{x}$ – 1) = x$^{2}$( 1 + x/2! + x$^{2}$/3! + x$^{3}$/4! + ….)

cos(x) = 1 – x$^{2}$/2! + x$^{4}$/4! – x$^{6}$/6! + x$^{8}$/8! – ….
=> (cos(x) – 1) = – x$^{2}$/2! + x$^{4}$/4! – x$^{6}$/6! + …. = –x$^{2}$(1/2! – x$^{2}$/4! + x$^{4}$/6! – ….)

$\lim_{x→0}$[x(e$^{x}$ – 1) + 2(cos(x) – 1)] / x(1 – cos(x))
= $\lim_{x→0}$[x$^{2}$( 1 + x/2! + x$^{2}$/3! + x$^{3}$/4! +….) + –2x$^{2}$(1/2! – x$^{2}$/4! + x$^{4}$/6! – ….)] / x$^{3}$(1/2! – x$^{2}$/4! + x$^{4}$/6! – ….)
= $\lim_{x→0}$[( 1 + x/2! + x$^{2}$/3! + x$^{3}$/4! +….) + –2(1/2! – x$^{2}$/4! + x$^{4}$/6! – ….)] / x(1/2! – x$^{2}$/4! + x$^{4}$/6! – ….)
= $\lim_{x→0}$[( 1 + x/2! + x$^{2}$/3! + x$^{3}$/4! +….) + (-1 + 2x$^{2}$/4! - 2x$^{4}$/6! + ….)] / x(1/2! – x$^{2}$/4! + x$^{4}$/6! – ….)
= $\lim_{x→0}$[(x/2! + x$^{2}$/3! + x$^{3}$/4! +….) + (2x$^{2}$/4! - 2x$^{4}$/6! + ….)] / x(1/2! – x$^{2}$/4! + x$^{4}$/6! – ….)
= $\lim_{x→0}$[(1/2! + x/3! + x$^{2}$/4! +….) + (2x/4! - 2x$^{3}$/6! + ….)] / (1/2! – x$^{2}$/4! + x$^{4}$/6! – ….)
= $\lim_{x→0}$[(1/2! + x/3! + x$^{2}$/4! +….) + (2x/4! - 2x$^{3}$/6! + ….)] / (1/2! – x$^{2}$/4! + x$^{4}$/6! – ….)
= $\lim_{x→0}$(1/2)/(1/2) = 1

Method 2 :- L'Hopital's Rule(apply only for →0/→0 or →$\infty$/→$\infty$ indeterminate form)

$\lim_{x→0}$[x(e$^{x}$ – 1) + 2(cos(x) – 1)] / x(1 – cos(x))
= $\lim_{x→0}$[(e$^{x}$ – 1) + xe$^{x}$ - 2sin(x)] / [(1-cos(x)) + x sin(x)]  (→0/→0) 
= $\lim_{x→0}$[2e$^{x}$ + xe$^{x}$ - 2cos(x)] / [2sin(x) + x cos(x)]  (→0/→0)  
= $\lim_{x→0}$[3e$^{x}$ + xe$^{x}$ + 2sin(x)] / [3cos(x) - x sin(x)]  (not →0/→0)  
= $\lim_{x→0}$(3)/(3) = 1

 

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