Q.296. Limit question solved by power series.

Question 296.
Find the value of lim (x→∞) [(x^10 + 3x^9)^1/10 - x].

Answer 296.
Binomial expansion
(a + b)^n
= a^n + nC1 a^(n-1)b^1 + nC2 a^(n-2) b^2 + .... + nCna^(n-n) b^n

This can also be used for the expansion of (x^10 + 3x^9)^(1/10) even though combinations are normally defined for integer n. But using it even for fractional n, this binomial series will go to infinity.

Thus,
(x^10 + 3x^9)^(1/10)
= (x^10)^1/10 + (1/10)C1 (x^10)^(1/10 - 1) (3x^9)^1
         + (1/10)C2 (x^10)^(1/10 - 2) (3x^9)^2 + ... to ∞
= x + (3/10) + 9(1/10)(1/10 - 1)/2! x^(-19) x^18 + .... to ∞
= x + (3/10) / x^(1/10) - (81/200) / x + ... to ∞.
=> (x^10 + 3x^9)^1/10 - x = 3/10 - (81/200) / x + ... to ∞
=> limit (x →∞) [(x^10 + 3x^9)^1/10 - x] = 3/10

[Note all terms after 3/10 will have a power of x in the denominator which will become 0
on taking limit x → ∞.]

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