Question

Evaluate limit : lim x tends to 0 cos8x-cos2x/cos12x-cos4x

Answer 1

∴$lim_{x\to 0}\frac{cos8x-cos2x}{cos12x-cos4x}$ $=\frac{3}{32}$

Step-by-step explanation:

L Hospital's Rule : L Hospital's rule is for finding the limit of a function.

The limit as x tends to a of a function of the form $\frac{f(x)}{g(x)}$ , when the limit of f and g at a are such that the value of f(a) and g(a) are 0 and 0 or, ∞ and ∞.

Given function is

$lim_{x\to 0}\frac{cos8x-cos2x}{cos12x-cos4x}$

Here f(x) = cos 8x-cos2x , g(x) = cos 12x-cos4x

f(0)= cos8.0-cos 2.0 =1-1=0 and g(0) = cos12.0 - cos 4.0=1-1=0

Therefore it is satisfy L Hospital's Rule. Then differentiate f(x) and g(x) with respect to x.

$=lim_{x \to 0}\frac{-8sin8x+2sin2x}{-12sin12x+4sin4x}$

Again putting x=0 in $\frac{-8sin8x+2sin2x}{-12sin12x+4sin4x}$ we get $\frac{-8sin8.0+2sin2.0}{-12sin12.0+4sin4.0} =\frac{0}{0}$

So, it satisfy the  L Hospital's Rule. Again differentiating the denominator and numerator.

$=lim_{x\to 0}\frac{-16 cos8x+4cos2x}{-144cos12x+16cos4x}$

Now putting x=0

$=\frac{-16 cos8.0+4cos2.0}{-144cos12.0+16cos4.0}$

$=\frac{-16+4}{-144+16}$

$=\frac{12}{128}$

$=\frac{3}{32}$

∴$lim_{x\to 0}\frac{cos8x-cos2x}{cos12x-cos4x}$ $=\frac{3}{32}$

Answer 2

Step-by-step explanation:

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