Question

for what value of k , the function f(x) = { (1-cos4x)/ 8x2 , x not eql to 0

{ k , x=0 is continuos at x=0?

Answer 1

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Answer 2

Answer:

Yes, the function will be continuous.

Step-by-step explanation:

In the function,

We have been provided that,

$f(x)=\frac{(1-cos4x)}{8x^{2}}$

where, x is not equal to 0.

Now,

For solving the given function we have to solve this using the trigonometric identity,

cos4x = cos²2x - sin²2x

also,

sin²2x + cos²2x = 1

So,

On putting these values in the function we get,

$f(x)=\frac{(1-cos4x)}{8x^{2}}= \frac{sin^{2}2x+cos^{2}2x-cos^{2}2x+sin^{2}2x}{8x^{2}}\\f(x)=\frac{sin^{2}2x}{4x^{2}}=\frac{sin2x}{2x}\times \frac{sin2x}{2x}$

Now, as the conditions are given for the limit on the function, we can say that,

$f(x)=lim_{x\rightarrow 0}(\frac{sin2x}{2x})\times lim_{x\rightarrow0}(\frac{sin2x}{2x})\\f(x)=1\times 1=1\\f(x)=1$

That's why the value of the function is a constant.

f(x) = 1 = constant

Therefore, the function will be continuous at, x = 0 as well.