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 Homework Statement:
 Suppose ##f## is differentiable at ##a##. Prove ##\lim_{h\rightarrow 0} \frac{f(a+h)  f(a)}{h} = f'(a)##
 Relevant Equations:

##f## is differentiable at ##a## means ##\lim_{x\rightarrow a} \frac{f(x)  f(a)}{xa}## exists and is finite.
##\lim_{x\rightarrow a} g(x) = L## means for any sequence ##(x_n)## that converges to ##a##, we have ##\lim_{n\rightarrow\infty} g(x_n) = L##.
Alternatively, ##\lim_{x\rightarrow a} g(x) = L## means for all ##\varepsilon > 0## there exists ##\delta = \delta(\varepsilon) > 0## such that ##\vert x  a \vert < \delta## implies ##\vert g(x)  L \vert < \varepsilon##.
Proof: By definition of derivative,
$$f'(a) = \lim_{x\rightarrow a}\frac{f(x)  f(a)}{x  a}$$
exists and is finite. Let ##(x_n)## be any sequence that converges to ##a##. By definition of limit, we have $$\lim_{x_n\rightarrow a} \frac{f(x_n)  f(a)}{x_n  a} = f'(a)$$. By definition of convergence, for all ##\varepsilon > 0## there exists ##N = N(\varepsilon) > 0## such that ##n > N## implies ##\vert x_n  a \vert < \varepsilon##, i.e. ##\lim_{n\rightarrow\infty} (x_n  a) = 0##. Let ##h = x_n  a##? Then
$$\lim_{h \rightarrow 0} \frac{f(a + h)  f(a)}{h} = f'(a)$$. []
I'm pretty sure this is wrong because I said ##h = x_n  a##, but ##h## is a constant and ##x_n  a## is a sequence.. also I don't think I can substitute ##h## under the limit like I did, but i'm not sure.
$$f'(a) = \lim_{x\rightarrow a}\frac{f(x)  f(a)}{x  a}$$
exists and is finite. Let ##(x_n)## be any sequence that converges to ##a##. By definition of limit, we have $$\lim_{x_n\rightarrow a} \frac{f(x_n)  f(a)}{x_n  a} = f'(a)$$. By definition of convergence, for all ##\varepsilon > 0## there exists ##N = N(\varepsilon) > 0## such that ##n > N## implies ##\vert x_n  a \vert < \varepsilon##, i.e. ##\lim_{n\rightarrow\infty} (x_n  a) = 0##. Let ##h = x_n  a##? Then
$$\lim_{h \rightarrow 0} \frac{f(a + h)  f(a)}{h} = f'(a)$$. []
I'm pretty sure this is wrong because I said ##h = x_n  a##, but ##h## is a constant and ##x_n  a## is a sequence.. also I don't think I can substitute ##h## under the limit like I did, but i'm not sure.