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$f(x)$ and $g(x)$ both differentiable twice at $x = a$ and we know that $f''(a) =g''(a)+f(a)$, $f(a) = g(a) = f'(a) = g'(a) \not = 0$

(we don't know if $f(x)$ and $g(x)$ are differentiable anywhere else except $x=a$)

prove: $\lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} = 1$

I tried to substitute $f^2(x)$ and $g^2(x)$ with a square of taylor expansion of degree 1 (with remainder of degree 2) for example:

$$f^2(x)= (f(a) + f'(a)x + \frac{f''(\xi)}{2}x^2)^2$$

same for $g^2(x)$.

Now because $f''(x)$ is continuous and since $x \rightarrow a$ we can conclude that $f(\xi) = f(a)$.

Now after simplifying the original expression i get:

$$\lim\limits_{x \rightarrow a} \frac{f^2(a)(3x^2+2x) + f(a)g''(a) \frac{x^4}{2}}{f^2(a) \cdot (x^2 + \frac{x^2}{2}) + f(a)(x+\frac{x^2}{2})g''(a)x^2 + g''}$$

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  • $\begingroup$ If $f$ and $g$ are twice differentiable at $x=a$, then $f'$ and $g'$ must exist in some neighborhood of $a$. But you don't know that $f''$ exists anywhere else (so you can't say it's continuous). $\endgroup$ – Barry Cipra May 3 '16 at 13:03
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Let's let $L\not=0$ denote the common value of $f(a)=g(a)=f'(a)=g'(a)$. Note that $f^2(x)-g^2(x)=(f(x)+g(x))(f(x)-g(x))$, so it suffices to show that

$$\lim_{x\to a}{f(x)-g(x)\over(f(x)-f(a))^2}={1\over2L}$$

since $\lim_{x\to a}(f(x)+g(x))=2L\not=0$. Invoking L'Hopital, it suffices to show

$$\lim_{x\to a}{f'(x)-g'(x)\over2f'(x)(f(x)-f(a))}={1\over2L}$$

which, since $2f'(x)\to2f'(a)=2L\not=0$, reduces to showing

$$\lim_{x\to a}{f'(x)-g'(x)\over f(x)-f(a)}=1$$

If we knew that $f$ and $g$ are both twice differentiable in a neighborhood of $x=a$ (and if the second derivatives are continuous at $a$), then we could invoke L'Hopital again. But it turns out we don't need to invoke L'Hopital at all; we can instead use the definition of derivative: Using the fact that $f'(a)=g'(a)$, we have

$${f'(x)-g'(x)\over f(x)-f(a)}={(f'(x)-f'(a))-(g'(x)-g'(a))\over f(x)-f(a)}={{f'(x)-f'(a)\over x-a}-{g'(x)-g'(a)\over x-a}\over {f(x)-f(a)\over x-a}}$$

hence

$$\lim_{x\to a}{f'(x)-g'(x)\over f(x)-f(a)}={\lim_{x\to a}{f'(x)-f'(a)\over x-a}-\lim_{x\to a}{g'(x)-g'(a)\over x-a}\over\lim_{x\to a} {f(x)-f(a)\over x-a}}={f''(a)-g''(a)\over f'(a)}={(g''(a)+L)-g''(a)\over L}=1$$

as desired.

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  • $\begingroup$ simple answer which assumes nothing beyond the hypotheses given in question. +1 $\endgroup$ – Paramanand Singh May 4 '16 at 6:51
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Since $f(a)=g(a)$, numerator en denominator tend to $0$ for $x \to a$. Apply l'Hôpital's rule: $$\lim_{x \to a} \frac{f^2(x)-g^2(x)}{(f(x)-f(a))^2} = \lim_{x \to a} \frac{2f(x)f'(x)-2g(x)g'(x)}{2(f(x)-f(a))f'(x)} = \lim_{x \to a} \frac{f(x)f'(x)-g(x)g'(x)}{f(x)f'(x)-f(a)f'(x)}$$ Since also $f(a)=g(a)=f'(a)=g'(a)$, you again have "$0/0$" so apply l'Hôpital's rule again: $$\begin{array}{rl} = & \displaystyle \lim_{x \to a} \frac{f(x)f''(x)+f'(x)^2-g(x)g''(x)-g'(x)^2}{f(x)f''(x)+f'(x)^2-f(a)f''(x)} \\[7pt] = & \frac{\displaystyle \lim_{x \to a} \left( f(x)f''(x)+f'(x)^2-g(x)g''(x)-g'(x)^2 \right) }{\displaystyle \lim_{x \to a} \left( f(x)f''(x)+f'(x)^2-f(a)f''(x)\right)} \quad \color{green}{(*)} \end{array} $$ Now with $\color{blue}{f''(a) = g''(a)+f(a)}$ and $\color{red}{f(a)=g(a)=g'(a)}$, the numerator becomes: $$f(a)\left( \color{blue}{g''(a)+f(a)} \right)+f'(a)^2-g(a)g''(a)-g'(a)^2 \\ = \color{red}{g(a)}\left( \cancel{\color{blue}{g''(a)}}+\bcancel{\color{red}{g(a)}} \right)+f'(a)^2\require{cancel}\cancel{-g(a)g''(a)}\bcancel{-\color{red}{g(a)^2}} \\ = f'(a)^2$$ and the denominator becomes: $$\require{cancel} \cancel{f(a)f''(a)}+f'(a)^2\cancel{-f(a)f''(a)} = f'(a)^2$$ So: $$\color{green}{(*)} = \frac{f'(a)^2}{f'(a)^2} = 1$$ Caveat: this method requires continuity of $f''$ at $a$, so that $\lim_{x \to a}f''(x) = f''(a)$.

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  • $\begingroup$ What if $f''(x)$ and $g''(x)$ exist only at $x=a$? $\endgroup$ – Barry Cipra May 3 '16 at 13:00
  • $\begingroup$ Someone else will have to come up with a solution for that case ;o). I added a warning thanks to your comment. $\endgroup$ – StackTD May 3 '16 at 13:01
  • $\begingroup$ I've posted an answer that doesn't assume continuity of the second derivative. $\endgroup$ – Barry Cipra May 3 '16 at 14:33
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If you use the Taylor polynomial, the answer is easy and short. Note $$ f(x)=f(a)+f'(a)(x-a)+\frac{1}{2}f''(\xi)(x-a)^2, g(x)=g(a)+g'(a)(x-a)+\frac{1}{2}g''(\eta)(x-a)^2$$ where $\xi$ and $\eta$ are between $a$ and $x$. So \begin{eqnarray} \lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} &=&\lim\limits_{x \rightarrow a} (f(x)+g(x))\frac{f(x)- g(x)}{(f(x) -f(a))^2}\\ &=&\lim\limits_{x \rightarrow a} 2f(a)\frac{\frac{1}{2}(f''(\xi)- g''(\eta))(x-a)^2}{[f'(a)(x-a)+O((x-a)^2)]^2}\\ &=&\lim\limits_{x \rightarrow a} f(a)\frac{f''(\xi)- g''(\eta)}{[f'(a)+O((x-a))]^2}\\ &=&f(a)\frac{f(a)}{f^2(a)}\\ &=&1. \end{eqnarray}

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  • $\begingroup$ we are only given the existence of $f''(a), g''(a)$. We don't know if $f''(\xi), g''(\eta)$ exist or not. $\endgroup$ – Paramanand Singh May 4 '16 at 6:50
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We can use Taylor series also (instead of L'Hospital's Rule). We have \begin{align} f(a + h) &= f(a) + hf'(a) + \frac{h^{2}}{2}f''(a) + o(h^{2})\notag\\ g(a + h) &= g(a) + hg'(a) + \frac{h^{2}}{2}g''(a) + o(h^{2})\notag \end{align} We can proceed as follows \begin{align} L &= \lim_{x \to a}\frac{f^{2}(x) - g^{2}(x)}{(f(x) - f(a))^{2}}\notag\\ &= \lim_{x \to a}\{f(x) + g(x)\}\cdot\frac{f(x) - g(x)}{(f(x) - f(a))^{2}}\notag\\ &= \{f(a) + g(a)\}\lim_{x \to a}\dfrac{f(x) - g(x)}{\left(\dfrac{f(x) - f(a)}{x - a}\right)^{2}(x - a)^{2}}\notag\\ &= \frac{2f(a)}{\{f'(a)\}^{2}}\lim_{x \to a}\frac{f(x) - g(x)}{(x - a)^{2}}\notag\\ &= \frac{2}{f(a)}\lim_{h \to 0}\frac{f(a + h) - g(a + h)}{h^{2}}\text{ (putting }x = a + h)\notag\\ &= \frac{2}{f(a)}\lim_{h \to 0}\frac{1}{h^{2}}\cdot\{f(a) + hf'(a) + \frac{h^{2}}{2}f''(a) - g(a) - hg'(a) - \frac{h^{2}}{2}g''(a) + o(h^{2})\}\notag\\ &= \frac{2}{f(a)}\cdot\frac{f''(a) - g''(a)}{2}\notag\\ &= \frac{2}{f(a)}\cdot\frac{f(a)}{2}\notag\\ &= 1\notag \end{align}

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