Let f(x)=#{tan(e^2)x^2-tan(-e^2)x^2}/sin^2 x#, x is not equal to zero, then the value of f(0) so that f is a continuous function is? 1)15 2)10 3)7 4)8

Generally, if we have a function #f# with is not continuous for some #c# (#c# is a discontinuity of #f#), we remove this discontinuity by defining a new function, #g#:

#g(x)={(f(x)" if " x!=c),(L " if " x=c) :}#

where #L = lim_(x->c) f(x)#. To see more on how to remove a removable discontinuity, see this answer by Jim H.

As such, we have to define #f# as

#f(x) = {((2tan(e^2)x^2)/sin^2x " if " x!=0),(L " if " x=0) :}#

Note: We didn't change anything about the upper part of the function. As the tangent function is odd, #tan(-e^2) = -tan(e^2)#.

So, we must find the value of #L = lim_(x->0) f(x)#.

#L=lim_(x->0) (2tan(e^2)x^2)/sin^2x#

As #2tan(e^2)# is a constant and not a variable in terms of #x#, we can take it out the front:

#L=2tan(e^2) lim_(x->0) x^2/sin^2x=2tan(e^2) lim_(x->0) (x/sinx)^2#

As the limit is linear,

#L = 2tan(e^2) (lim_(x->0) x/sinx)^2#

This is basic limit:

#lim_(x->0) x/sinx =1#

#:. L = 2tan(e^2)1^2=2tan(e^2)#

Which is none of the answers, however, by letting

#f(x) = {((2tan(e^2)x^2)/sin^2x " if " x!=0),(2tan(e^2) " if " x=0) :}#

does remove the discontinuity.