The limit of limits!

Calculus Level 5

Given f ( x ) f(x) is a polynomial of n n th degree and

lim ⁡ x → 0 2 r − 1 ( x − 2 ) f ( x ) − x r + 2 r x r ( x − 2 ) = 1 2 ∀ r ∈ N ∩ [ 1 , n ] \large \lim_{x \to 0} \frac{2^{r-1}(x-2)f(x)-x^{r}+2^{r}}{x^{r}(x-2)}=\frac{1}{2} \quad \forall r \in \mathbb N \cap [1,n]

Find lim ⁡ n → ∞ f ( 1 ) \displaystyle \lim_{n \to \infty}f(1) .

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The answer is 2.00.

Rohith M.Athreya by Rohith M.Athreya , 22, India

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1 solution

Brian Moehring
Mar 7, 2017

Let r = n r=n in the given limit and re-write the limit as lim ⁡ x → 0 f n ( x ) − 1 2 n − 1 x n − 2 n x − 2 x n = 1 2 n \lim_{x\rightarrow 0} \frac{f_n(x) - \frac{1}{2^{n-1}}\frac{x^n - 2^n}{x-2}}{x^n} = \frac{1}{2^n} and since 1 2 n − 1 x n − 2 n x − 2 \frac{1}{2^{n-1}}\frac{x^n-2^n}{x-2} is a polynomial of degree n − 1 ≤ n n-1 \leq n and f n ( x ) f_n(x) is a polynomial of degree n n , this limit implies f n ( x ) = 1 2 n x n + 1 2 n − 1 x n − 2 n x − 2 f_n(x) = \frac{1}{2^n}x^n + \frac{1}{2^{n-1}}\frac{x^n - 2^n}{x-2}

Setting x = 1 x=1 yields f n ( 1 ) = 1 2 n + 1 2 n − 1 1 − 2 n 1 − 2 = 1 2 n + 2 − 1 2 n − 1 f_n(1) = \frac{1}{2^n} + \frac{1}{2^{n-1}}\frac{1-2^n}{1-2} = \frac{1}{2^n} + 2 - \frac{1}{2^{n-1}} and then taking the limit lim ⁡ n → ∞ f n ( 1 ) = 0 + 2 + 0 = 2 . \lim_{n\rightarrow\infty} f_n(1) = 0 + 2 + 0 = \boxed{2}.

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