Which type of discontinuity exists at x=0?

removable

Step-by-step explanation:

f(x) = cos(1/x^2) at x = 0;

For the function f(x) = cos(1/x^2) let's substitute the value of 0 in for x...

= cos(1/(0)^2)

= cos(1/0)

Now remember that x/0 is undefined. Removable discontinuity is undefined at the point x₀, and hence our solution is "removable."

assuming you mean addition: addition and subtraction are reverse operations.

step-by-step explanation:

24

step-by-step explanation: