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Given $g$ is differentiable at $x = a$, prove the following limit exists

  • $\lim_{h \to 0} \frac{g(a+6h)-g(a-6h)}{12h}$

After reading this similar post:

if $f$ is differentiable at $x_0$ then the limit exists

I tried applying similar logic but I'm stuck at this point because I don't know how to deal with $6h$ and $12h$.

  • $\lim_{h \to 0} \frac{g(a+6h)-g(a-6h)}{12h}$ = $\lim_{h \to 0} \frac{g(a+6h) -g(a) + g(a) - g(a-6h)}{12h}$

I imagine you can set $f(x) = g(a+6h)$ but this is where I got up to. Any tips?

Edit: If $\lim_{h \to 0} \frac{g(a+h) -g(a)}{h}$ = $\lim_{h \to 0} \frac{g(a+6h) -g(a)}{12h}$ then I see how I can finish the question

So,

  • $\lim_{h \to 0} \frac{1}{2}\frac{g(a+6h) -g(a) + g(a) - g(a-6h)}{6h}$ = $\frac{1}{2}\lim_{h \to 0} \frac{g(a+6h) -g(a) + g(a) - g(a-6h)}{6h}$ = $\frac{1}{2}$$\lim_{h \to 0} (\frac{g(a+6h) -g(a)}{6h} + \frac{g(a) - g(a-6h)}{6h})$ = $\frac{1}{2}$$\lim_{h \to 0} (\frac{g(a+6h) -g(a)}{6h} + \frac{g(a+6h^*) - g(a)}{6h^*})$ = $\frac{1}{2}$$\lim_{h \to 0} \frac{g(a+6h) -g(a)}{6h}$ + $\frac{1}{2}$$\lim_{h \to 0} \frac{g(a+6h^*) -g(a)}{6h^*}$

And we know that since $\lim_{h \to 0} \frac{g(a+h) -g(a)}{h}$ exists and is equal to $\lim_{h \to 0} \frac{g(a+6h) -g(a)}{6h}$ we have

  • $\frac{1}{2}$$\lim_{h \to 0} \frac{g(a+6h) -g(a)}{6h}$ + $\frac{1}{2}$$\lim_{h \to 0} \frac{g(a+6h^*) -g(a)}{6h^*}$ that also exist
Tingo Hugo
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2 Answers2

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As a hint : use a substitution $\lim_{h \to 0} \frac{g(a+6h)-g(a-6h)}{12h} $ for example use $6h =u \to h=\frac {1}{6}u$ so $$\lim_{h \to 0} \frac{g(a+6h)-g(a-6h)}{12h}=\\\lim_{u \to 0} \frac{g(a+u)-g(a-u)}{2u}\\\lim_{u \to 0} \frac{g(a+u)-g(a)+g(a)-g(a-u)}{2u} $$Hope it helps you

Khosrotash
  • 24,922
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By definition, using little-o notation, we have

  • $g(a+6h)=g(a)+g'(a)\cdot 6h + o(h)$
  • $g(a-6h)=g(a)+g'(a)\cdot (-6h) + o(h)$

and then

$$\frac{g(a+6h)-g(a-6h)}{12h}=g'(a)+o(1)\to g'(a)$$

Refer also to:

user
  • 154,566
  • Where did the $o(h)$ come from? Why is it not just $g(a) + g'(a)6h$? – Tingo Hugo Feb 06 '24 at 14:45
  • @TingoHugo It is an equivalent definition of differentiability. Refer also to https://math.stackexchange.com/q/4061952/505767 – user Feb 06 '24 at 14:51
  • @TingoHugo $o(h)$ is a remainder term it is zero if the function is locally a line otherwise we have also this term. – user Feb 06 '24 at 14:54
  • Oh cool, I've never come across that but thats interesting – Tingo Hugo Feb 06 '24 at 15:00
  • This is an important concept indeed and very useful as a first (order) taste for the more general Taylor expansion. – user Feb 06 '24 at 15:04