discuss the continuity of f(x) = lim n–>infinity ( x^2n-1) / (x^2n–1) for real x
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Answer:
Correct option is
A
f(x)=1 for ∣x∣>1
B
f(x)=−1 for ∣x∣<1
Let us take ∣x∣<1, then x=
q
1
∴x
2n
=(
q
1
)
2n
(
q
1
)
2n
→0 when n→∞
Substituting this value in the limit we get,
f(x)=
0+1
0−1
=−1
Again, let us take ∣x∣>1 then we can write the limit as
n⟶∞
lim
x
2n
x
2n
(
1+
x
2n
1
1−
x
2n
1
)
=
1+0
1−0
=1
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