Question

We have `f:RR->RR,f(x)=|x|root(3)(1-x^2)`.Is this function continous in `x=0`?

Answer 1

Yes. See the explanation.

Explanation:

Function is continous at the point if it is defined at the point and if exist left and right limits and their values are equal to the value of the function in the point. So,

`f(0) = |0|root(3)(1-0^2) = 0`

Let `epsilon in R^+, epsilon->0`, then:

`L = lim_(x->0-epsilon) |x|root(3)(1-x^2) = lim_(epsilon->0)|-epsilon|root(3)(1-(-epsilon)^2)`

`L = lim_(epsilon->0) epsilon root(3)(1-epsilon^2) = 0`

`R = lim_(x->0+epsilon) |x|root(3)(1-x^2) = lim_(epsilon->0)|epsilon|root(3)(1-epsilon^2) = 0`

Obviously, `L=f(0)=R` and function is contious in `x=0`.