Question

Lim x tends to 0 [ { (1-cos2x) (3+cosx)} / (xtan4x)} =?

Answer 1

Lim x tends to 0 [ { (1-cos2x) (3+cosx)} / (xtan4x)} 
=Lim x tends to 0 [ { 2(sin^2x) (3+cosx)} / (xtan4x)} 
=2 Lim x tends to 0 [ { (sinx*sinx) (3+cos0)} / (xtan4x)} 

=2*4 Lim x tends to 0 [ { (sinx*sinx) } / (xtan4x)} =8*1*1/4=2

Answer 2

The value of the given limit is 2.

We have to find the value of ,

$\lim_{x \to 0} \frac{(1-cos2x)(3+cosx)}{xtan4x}$

At first we should solve the given expression, [(1 - cos2x)(3 + cosx)]/(xtan4x)

[we know, (1 - cos2Ф) = 2sin²Ф ]

$\frac{(1-cos2x)(3+cosx)}{xtan4x}\\\\=\frac{(2sin^2x)(3+cosx)}{xtan4x}$

$Now, \lim_{x \to 0} \frac{(2sin^2x)(3+cosx)}{xtan4x}\\\\=2(3+cos0^{\circ})\lim_{x\to 0}\frac{\left(\frac{sinx}{x}\right)sinx}{tan4x}\\\\=8\lim_{x\to 0}\frac{sinx}{x}.\frac{\frac{sinx}{x}}{\frac{tan4x}{4x}}.\frac{x}{4x}\\\\=2$

Therefore the value of given limit is 2.

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