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Let $f(x)$ be in $C^2 $ at $x=2$ and $f ' (2) =-1 $ . Let $a_n$ be a sequence such that $a_n\neq 2 $ for every $n$ and $lim_{n \to \infty} a_n =2 $ .

Define:

$b_n = \frac{f(a_n)-f(2)}{a_n-2} $ .

Calculate the following limit using $f''(2)$:

$lim_{n\to \infty} \frac{b_n+1 }{a_n-2 }$ (Hint: use Taylor expansion of order one)

My attempt: 1) All I know is, that $b_n \to f'(2) =-1 $ . I can't understand how to use a Taylor expansion of order 1 in order to solve this

Please help me. I'm completely lost !

Thanks in advance!

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  • $\begingroup$ In a Taylor expansion of order one of $f$ around $2$, what is the remainder term in the Lagrangian form? $\endgroup$ – Daniel Fischer Jan 20 '14 at 19:47
  • $\begingroup$ @DanielFischer : we have that $R_1(x) = \frac{f^{(2)}(c)}{2} x^2 $ when $0 < c < x $ . Does it help ? Will you please help me understand how to solve this ? Thanks $\endgroup$ – homogenity Jan 20 '14 at 19:58
  • $\begingroup$ You have written it as if the Taylor expansion was about $0$, that may obscure it a bit. Expand about $2$ to get $$f(x) = f(2) + f'(2)\cdot (x-2) + R_1(x).$$ Look again at the remainder term, and then set $x = a_n$. $\endgroup$ – Daniel Fischer Jan 20 '14 at 20:02
  • $\begingroup$ OK. So: $ f(x) = f(2) + f'(2)\cdot (x-2) $, $ R_1(x)= \frac{f^{(2)}(c)}{2} (x-2)^2 $ when $c$ is between $2$ and $a_n$ . Putting $x=a_n$ , we get: $\endgroup$ – homogenity Jan 20 '14 at 20:06
  • $\begingroup$ $ f(a_n) = f(2) - (a_n -2) + \frac{f^{(2)}(c)}{2} (a_n-2)^2 $ . So: $b_n = - \frac{f^{(2)}(c)}{2} (a_n -2 ) $ , right? $\endgroup$ – homogenity Jan 20 '14 at 20:08
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f(a(n)) = f(2) + f'(2)(a(n) - 2) + 0.5f''(c)(a(n) - 2)^2 with 2 < c < a(n). So b(n) = f'(2) + 0.5f''(c)(a(n) - 2). So b(n) + 1 = f''(c)(a(n) - 2) and then (b(n) + 1)/(a(n) - 2) = 0.5f''(c) --> f''(2) when n --> infinity. So the answer is 0.5f''(2).

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