# Taylor Expansion -Question

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!

• In a Taylor expansion of order one of $f$ around $2$, what is the remainder term in the Lagrangian form? – Daniel Fischer Jan 20 '14 at 19:47
• @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 – homogenity Jan 20 '14 at 19:58
• 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 $x = a_n$. – Daniel Fischer Jan 20 '14 at 20:02
• 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: – homogenity Jan 20 '14 at 20:06
• $f(a_n) = f(2) - (a_n -2) + \frac{f^{(2)}(c)}{2} (a_n-2)^2$ . So: $b_n = - \frac{f^{(2)}(c)}{2} (a_n -2 )$ , right? – homogenity Jan 20 '14 at 20:08

## 1 Answer

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).