Cpt qa doubt - limits, differentiation.

3. taking logarithms on both sides we get log(f(x)) = (a+1+2x)(log(a+x) -log(1+x))

differentiating on both sides we get df/dx(1/f(x)) = 2*(log(a+x) - log(1+x)) + (a+1+2x) (1/(a+x) - 1/(1+x) )

df/dx = (2*(log((a+x)/(1+x)))+(a+1+2x)(1-a)/((a+x)*(1+x))) * ((a+x)/(1+x))^(a+1+2x) is the answer

2. substitute (1+x)^2 =t as x tends to 0 t tends to 1 now the limit becomes (t^3 -1)/(t-1) = limit t tends to 1 (t^2 + t + 1) =3

f(x) = (1+x)^6/(1+x)^2-1 = (1+x)^6*(1+x)^2+1/(1+x)^2-1)*(1+x)^2+1=(1+x)^6*(1+x)^2+1/(1+x)^2-(1)^2=(1+x)^6+1/-1

limit x tends to 0 = (1+0)^6+1/-1=1+1/-1=-2.

2.f(x) = (a+x/1+x)^a+1+2x

d/dx fx=d/dx (a+1+2x)log(a-1)

           =(a+1+2x)d/dxlog(a-1)+log(a-1)d/dx(a+1+2x)

           =(a+1+2x)*1/a-1+log(a-1)*(0+0+2)

           =2log(a-1)ans...

use lhospital rule which is as follows

check whether denominator tends to zero so that the entire expression tends to infinity

here denominator tends to zero (1+0)^2-1 = 1-1 = 0

By lhospital rule differentiate numerator and denominator separately

Differentiating numerator we get 6(1+x)^5 and differentiating denominator we get 2(1+x)

Now apply lt x->0     (6(1+x)^5)/2(1+x)   =   6(1)^5 / 2  =   6/2  =  3..

If again on applying limit the denominator tends to zero apply this theorem until the denominator bears a value other than 0...This method will be so useful in exam as it saves time considerably.

lhospital's rule applies when the expression is 0/0 not when the denominator is tending to infinity. if the 2nd question is posted correctly then the answer is limit does not exist as if x tends to 0- we get -infinity and if x tends to 0+ we get +infinity I assumed when answering the question that she forgot to include -1 in the numerator of the question

Thanks for all the answers.

The limit question is correct. 1 is not missed out. The question doesn't have -1 in the numerator...

then the answer is limit does not exist as if x tends to 0 from negative side the equation goes to -infinity and if x tends to 0 from positive side equation goes to +infinity