29what is the exact question .\frac{sinx-(sinx)^{sinx}}{1-sinx + lnx}
or
\frac{sinx - sin(x)^{sinx}}{1- sinx + lnx}
For the first one i think the answer is 0..
dunno abt 2nd...
12 Answers
29
1just place the value of x in the sum and you will get the ans my friend
it will come as 0/(a finite value)=0
1dude just substitue pi/2 in place of x ul get it as o .. numerator becomes 0 divided by a real no. which is zero.....
1sorry i made a typing mistake the ques is lim x→pie/2 (sinx - sinxsinx)/(1-sinx+lnsinx).....
1L -- hospital gives ,
lim_\; {x\rightarrow \pi / 2}\frac{cosx\;-\;[\;{sinx^{sinx }\;cosx\;+\;sinx^{sinx}\;{\;log\;sinx\;} \;cosx}\;]}{1\;-\;cosx\;+\;cotx}\;}
Putting x = pi/2 yields the limit to be 0 / 1 or simply 0.
1@ Michael --> u made a mistake while differentiating the denominator.....
1 will not be there (differentiation of a constant)
1is the answer infinity?????????
i got it by assuming sin x=t and putting lim t→1
1Michael actually there will be no 1 in the denominator.....Guys I have figured out the solution and the ans is 3..