13no idea bhai...post ur working naa...
If ...
\lim_{x \rightarrow \infty} \frac{ a \left(2x^3 -x^2 \right) +b \left(x^3+5x^2-1 \right) -c \left(3x^3+x^2 \right)}{a\left(5x^4-x \right)-bx^4 +c \left(4x^4+1 \right) +2x^2 + 5x} = 1,
then the value of (a+b+c) can be expressed in the lowest form as p/q. The value of (p+q) is ____ ?
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5 Answers
21\lim_{x\rightarrow \infty } \frac{x^{3}(2a+b-3c) + x^{2}(5b-a-c)-b}{x^{4}(5a-b+4c)+x(5-a)+x^{2}(2) +c}=1
now dividing numerator and denominator by x2
\lim_{x\rightarrow \infty } \frac{x(2a+b-3c) +(5b-a-c)-b/x^{2}}{x^{2}(5a-b+4c)+(5-a)/x +(2) +c/x^{2}}=1
now since the limit is finite so we must hav
2a+b-3c=0.....................i
and 5a-b+4c=0...............ii
and we know tat since x→∞
so -b/x2 →0
(5-a)/x→0
c/x2→0
and (5b-a-c)/2 =1.................iii
solve eq i ,ii,and iii
13Thnx fr the approach...am getting 107, see if ny error..
[EDIT : got my error; trying again :(]
19the method looks fine,,,that is what to be done ! i have not checked the calculations,though!