f(x) = {[(1-cos 2x)/2x^2,x<0][k,x=0][x/|x|,x>0

Question

If the function f(x), defined below is continuous at x = 0, find the value of k :

\(f(x) = \begin{cases}\frac{1-cos\,2x}{2x^2}&,\quad x <0\\k&,\quad x=0\\\frac{x}{|x|}&;\quad x>0\end{cases} \)

Answer 1

We have to find the value of 'k' 

Given : 

f(x) is continuous at x = 0 & f(0) = k 

If f(x) is be continuous at x = 0,

Then, 

f(0)^– = f(0)⁺ = f(0)

Since, 

f(x) is continuous at x = 0 & f(0) = k 

And also, 

f(0)^– = f(0)⁺ = f(0) 

So, 

k = 1