Limit X tends to 0

8/X^8(1-CosX^2/2-CosX^2/4+Cos X^2.CosX^2/4)

Open in App

Solution

Limit_x---->08/x⁸[(1-cosx²/2)(1-cosx²/4)]=Limit_x--->08[(1-cosx²/2)/x⁴][(1-cosx²/4)/x⁴]

=Limit_x---->08[2(sinx²/4)²/x⁴][2(sinx²/8)²/x⁴]=Limit_x---->032[(sinx²/4)/4(x²/4)]²[(sinx²/8)/8(x²/8)]²

=Limit_x--->032/(4²8²) [(sinx²/4)/(x²/4)]²[(sinx²/8)/(x²/8)]² =1/32*Limit_x---->0[(sinx²/4)/(x²/4)]²[(sinx²/8)/(x²/8)]²

=1/32

Here i used following formulas:

1-cosx=2sin²(x/2)

Limit_y--->0siny/y = 1 [replace y by x²/4 and x²/8 to get a new result which is used in solving above problem...]

Suggest Corrections

Similar questions

L t n \to \infty {c o s \frac{x}{2} . c o s \frac{x}{4} . c o s \frac{x}{8} \dots \dots c o s \frac{x}{2^{n}}}

cos \frac{x}{2} . cos \frac{x}{4} . cos \frac{x}{8} . . . . = \frac{sin x}{x}

\frac{1}{2^{2}} {sec}^{2} \frac{x}{2} + \frac{1}{2^{4}} {sec}^{2} \frac{x}{4} + . . . . = {csc}^{2} x - \frac{1}{x^{2}}

{tan}^{- 1} (\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}}) = \frac{π}{4} - \frac{x}{2}

π < x < \frac{3 π}{2}

{tan}^{- 1} (\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}}) = \frac{π}{4} - \frac{x}{2}

π < x < \frac{3 π}{2}

t a n^{- 1} (\frac{\sqrt{1 + c o s x} + \sqrt{1 - c o s x}}{\sqrt{1 + c o s x} - \sqrt{1 - c o s x}}) = \frac{π}{4} - \frac{x}{2}, where π < x < \frac{3 π}{2}

Join BYJU'S Learning Program