priya (Student) (64 Points)
25 May 2012Please help me with this question..Thanks in advance.
1) (1+x)^6 then lim f(x) is equal to is?
----------- x->0
(1+x)^2-1
2) If f(x) = {(a+x)/(1+x)}^(a+1+2x) the value of d [f(x)]
---
dx
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 25 May 2012
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
intermediate(ipc)course
(no)
(1460 Points)
Replied 25 May 2012
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.
intermediate(ipc)course
(no)
(1460 Points)
Replied 25 May 2012
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...
CA S.Parthasarathy
(Chartered Accountant)
(144 Points)
Replied 26 May 2012
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.
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 26 May 2012
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
priya
(Student)
(64 Points)
Replied 26 May 2012
Thanks for all the answers.
The limit question is correct. 1 is not missed out. The question doesn't have -1 in the numerator...
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 26 May 2012
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
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