f(x)=(4(with power x)-1)(with whole power 3)/ (sine of(x/4))* log(1+x(power2)/3)

is contonuous at x=0?

yes it is continues

f(x)=       (4^x-1)³

      ----------------------------

       sin(x/4) *  log(1+x^2/3)

see that by putting x=0 it gives 0/0 form. so  U CAN APPLY "L HOSPITAL RULE"(differentiating numerator and denominator independently)

now it is,

         3* (4^x-1)² * log4 * 4^x

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cos(x/4)*log(1+x^2/3)+ sin(x/4)*2/3*1/(x)^1/3*1/(1+x^2/3)

again 0/0 form by putting x=0...do the same thing as earlier.

it may be long bt this is the concept...keep on differentaing till we get 0/0 form or infinity/infinity form.

other option may be by expanding the gaven things using binomial expansion.

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