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

1 Answer
Apr 23, 2017

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#.