Proving the derivative of a certain point using the sequence definition

Using this derivative definition

If $f$ is a function and has derivative $f'(c)$ at the point $c$ in the domain of $f$ means that if ($a_n$)$_{n=1}^{\infty}$ is any sequence converging to $c$ such that $a_n$ $\not= c$is in the domain of $f$ for all $n \in \mathbb{N},$ then: $$(\,\frac{f(x_n)-f(c)}{x_n-c}) \,_{n=1}^{\infty}$$converges to $f'(c)$

prove that $f'(c)=-6,$ where $f(x)=x^2 +2$

I feel like this should be fairly straight forward, but I'm having trouble.

My attempt:

Plugging in $a_n$ for $x,$ and using $f(c)=f(-3)=11,$

$(\,\frac{f(x_n)-f(c)}{x_n-c}) \,_{n=1}^{\infty}$<=> $\frac{[(a_n)^2+2]-11}{a_n-3}$<=>$\frac{(a_n)^2-9}{a_n-3}$<=>$a_n-3$

Because we are using sequences, I'm not sure if this is a correct approach, and I am not sure how to complete the proof. Do I need to choose a certain epsilon to show that this sequence converges to $f'(c)$?

• So you have two convergences in this problem: the sequence $\{a_n\}\to c$ and $\frac{f(x)-f(c)}{x-c}\to f'(c)$ (as $x->c$) you can thus bound $|a_n-c|<\epsilon_1$ for $n>N_1$ and $|\frac{f(x)-f(c)}{x-c}-f'(c)|<\epsilon_2$ for $|x-c|<\delta$. Can you see how to combine these bounds to get a proof that the difference quotient sequence converges? Apr 27, 2017 at 17:18
• Are you allowed to use the fact that $\lim_{n \to c}(a_n + b_n) = \lim_{n \to c}a_n + \lim_{n \to c}b_n$ Apr 27, 2017 at 17:34
• @wgrenard Yes, I have already proven that theorem! Apr 27, 2017 at 17:46
• @Twis7ed So that would involve a proof using the sequential definition of limits and the delta-epsilon definition? Apr 27, 2017 at 17:50

What you have done is fine, and you don't have to get into any business with $\epsilon$ because you already know that $\lim_{n \to \infty}a_n = -3$
Since $\lim_{n \to \infty}a_n$ exists and is finite, and clearly $\lim_{n \to \infty} 3 = 3$ so it also exists and is finite, you can conclude that
$$\lim_{n \to \infty}(a_n - 3) = \left( \lim_{n \to \infty}a_n \right) - \left ( \lim_{n \to \infty} 3 \right) = -3-3 = -6$$
• No worries! And we know $a_n$ converges to -3 from the definition I gave? Apr 28, 2017 at 3:02
• @Mathgirl correct. It is assumed that $a_n$ is a sequence converging to $c$. Apr 28, 2017 at 3:27