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The Question:

Verify that if $\lim_{x\to a}f(x)$ exists, then $\lim_{x\to a}f(x)=\lim_{h\to 0}f(a+h)$.

My plan is to use the theorem showing that the limit of a sum is the sum of the limits. By that theorem, $\lim_{h\to 0}f(a+h)=\lim_{h\to 0}f(a)+\lim_{h\to 0}f(h)=f(a)+0=f(a)=\lim_{x\to a}f(x)$.

Does this proof hold? Please let me know if there is anything that I need to clarify.

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  • $\begingroup$ Can you state clearly your theorem? It seems to me that the theorem cannot be applied in your case. $\endgroup$ – user99914 Oct 21 '13 at 3:42
  • $\begingroup$ Generally, if lim(x->a)f(x) exists, then lim(x->a)f(x+h) = lim(x->a)f(x)+lim(x->a)f(h). Is this incorrect? $\endgroup$ – Heath Huffman Oct 21 '13 at 3:45
  • $\begingroup$ you can't write $\lim_{h \to 0}f(a + h) = \lim_{h \to 0}f(a) + \lim_{h \to 0}f(h)$ $\endgroup$ – Paramanand Singh Oct 21 '13 at 3:45
  • $\begingroup$ I think the theorem you mentioned is $\lim_{x\to a} (g(x) + f(x)) = \lim_{x\to a} g(x) + \lim_{x\to a}f(x)$. It involves two functions $f$ and $g$. In your case you have only one function. $\endgroup$ – user99914 Oct 21 '13 at 3:47
  • $\begingroup$ Okay I see now. So which direction should I move in to prove this? $\endgroup$ – Heath Huffman Oct 21 '13 at 3:59
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The limit $\lim_{x\to a}f(x)$ exists and is equal to $b$: $$\lim_{x\to a}f(x)=b$$ if and only if for each $\varepsilon\in\mathbb R^+$ exists $\delta\in\mathbb R^+$ such that $$\forall x,0<|x-a|<\delta\implies|f(x)-b|<\varepsilon.$$

Hint: Apply analogous definition for $\lim_{h\to0}f(a+h)$ and tell if you can derive one from the other.

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  • $\begingroup$ "such that" (FYI) $\endgroup$ – The Chaz 2.0 Oct 21 '13 at 4:09
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Well first we have to prove that differentiability implies continuity. (We have to prove that $\lim_{x\to a}f(x)$ exists.)

Thus, suppose we have a differentiable function $f(x)$ such that $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$. This equals $f'(a)$.

Now, we multiply each side by $h$. Thus, we get $\lim_{h\to 0}hf'(a)$. (What is the answer here?)

Thus, we get $\lim_{h\to 0}f(a+h) \to f(a)$. (Why?)

Thus, we set $x = a + h$. Thus, $\lim_{x-a \to 0}f(x) \to f(a)$.

Thus, we have proven that such a $\lim_{x\to a}f(x) = f(a)$ exists.

The answer to your question:

To answer your question, go backwards!

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  • $\begingroup$ "Well first we have to prove that differentiability implies continuity." Why? $\endgroup$ – Pedro Tamaroff Oct 21 '13 at 4:15
  • $\begingroup$ "Verify that $lim_{x\to a}f(x)$ exists" $\endgroup$ – Don Larynx Oct 21 '13 at 4:20
  • $\begingroup$ What does that have to do with differentiability? We're talking continuity here. $\endgroup$ – Pedro Tamaroff Oct 21 '13 at 4:21
  • $\begingroup$ I agree with Pedro Tamaroff, this is a simple question on limits, why do we need to invoke differentiation? $\endgroup$ – Paramanand Singh Oct 21 '13 at 4:31
  • $\begingroup$ It was just another way of rephrasing the question! Hence the parentheses. Isn't it still correct? $\endgroup$ – Don Larynx Oct 21 '13 at 4:38

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