Question

as limit x tends to zero , what is the value of tan^4x - sin^4x whole divided by x^6

Answer 1

limx→04x−4sinx4x−tan4x=−12

Answer 2

Answer:

The value of given limit is

$L = lim_{x - > 0} \frac{tan {}^{4}x - sin {}^{4} x }{x {}^{6} } = 2$

Explanation:

The given limit is

$L = lim_{x - > 0} \frac{tan {}^{4}x - sin {}^{4} x }{x {}^{6} }$

It can be simplified as follows

$= lim_{x - > 0} \frac{ \frac{sin {}^{4}x }{cos{}^{4}x } -sin {}^{4} x }{x {}^{6} } \\ = lim_{x - > 0} \frac{sin {}^{4}x(1 - cos {}^{4} x)}{cos {}^{4}x.x {}^{6} }$

This limit can be further simplified using the following relations given below

$lim_{x - > 0} \frac{1}{cos {}^{4} x} = 1 \\ 1 - cos {}^{4} x = (1 - cos {}^{2} x)(1 + cos {}^{2} x) \\ = sin {}^{2} x(1 + cos {}^{2} x)$

Therefore,the above limit can be written as follows

$L = lim_{x - > 0} \frac{sin {}^{4} x.sin {}^{2} x(1 + cos {}^{2}x) }{1 \times x {}^{6} } \\ = lim _{x - > 0} \frac{sin {}^{6}x(1 + cos {}^{2}x) }{x {}^{6} } \\ = lim_{x - > 0} \frac{sin {}^{6} x(1 + 1)}{x {}^{6} } \\ = 2lim_{x - > 0} (\frac{sinx}{x} ) {}^{6}$

From the knowledge of limits we know a fundamental thing i.e

$lim_{x - > 0} \frac{sinx}{x} = 1$

Therefore substituting this in given limit,we get our limit as

$L = 2 \times 1 {}^{6} = 2$

Therefore the value of given limit is

$L = lim_{x - > 0} \frac{tan {}^{4}x - sin {}^{4} x }{x {}^{6} } = 2$

#SPJ2