UPSC NDA I 2015 Mathematics Solved Paper

Consider the following transformation
x=

1

n

x→∞,n→0
f(x)=

2+ 1 n2

1+ 1 n

−

A

n

−B
=

2n2+1

n2+n

−

A

n

−B
=

2n2+1−A(n+1)−B(n2+n)

n2+n

=

n2(2−B)−n(A+B)+1−A

n2+n

Now as ‘n’ tends to 0 the numerator and denominator, both should yield 0 to obtain an indefinite form. Hence for this to happen, the value of A has to 1.
f(x)=

n2(2−B)−n(1+B)

n2+n

lim

n→0

f(x)
=

lim

n→0

[

n2(2−B)−n(1+B)

n2+n

]
=

lim

n→0

[

2n(2−B)−(1+B)

2n+1

] (Using L Hopital's rule )
=

−(1+B)

1

=3
B+1=−3
B=−4