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I'm currently taking an introduction to Calculus course and I've come across the following identity:

How would one come up with this? My best guess is using L'Hospital's Rule on $$\lim_{x\rightarrow a}{\frac{f(x)-f(a)}{x-a}}$$

but I'm not very sure how, since differentiating both the numerator and denominator merely yields

$$\lim_{x\rightarrow a}{f'(x)} = f'(a)$$

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The result holds under the weaker assumption that $f''(a) $ exists (other answers assume the continuity of $f''$ at $a$ or even more). Also note that under this weaker assumption it is not possible to apply L'Hospital's Rule on the expression under limit in question and hence a slight modification is required.


By definition of derivative we have $$\lim_{x\to a} \frac{f'(x) - f'(a)} {x-a} =f''(a)\tag{1}$$ Adding this to the limit in question it is clear that our job is done if we can establish that $$\lim_{x\to a} \frac{f(x) - f(a) - (x-a) f'(a)} {(x-a)^2}=\frac{f''(a)}{2}\tag{2}$$ And the above limit is easily evaluated by a single application of L'Hospital's Rule. Applying it on the fraction on left side we get a new fraction $$\frac{f'(x) - f'(a)} {2(x-a)}$$ which clearly tends to $f''(a) /2$ (via $(1)$) and hence the fraction on left side of $(2)$ also tends to the same value and the identity $(2)$ is established.

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Hint:

Set $g(x)=f(x)-f(a) -(x-a)f'(x)$ and apply L'Hospital's rule twice:

  • $g'(x)= f'(x)-f'(x)-(x-a)f''(x)=-(x-a)f''(x)$,
  • $g''(x)= -f''(x)-(x-a)f'''(x)$.
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$$\frac{d}{dx}(\textbf{numerator}) =f’(x)-(f’(x)+f’’(x)(x-a))=-f’’(x)(x-a).$$

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No, you're just applying L'Hopital's rule to the limit on the LHS. Since $f(a)$ is a constant, by the product rule the limit evaluates to

$$\lim\limits_{x\to a}\frac{f'(x)-((x-a)f''(x)+f'(x))}{2(x-a)}=\lim\limits_{x\to a}\frac{-(x-a)f''(x)}{2(x-a)}=\lim\limits_{x\to a}-\frac{f''(x)}{2}$$

Evaluating that limit at $a$ gives you the derired answer.

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  • $\begingroup$ Actually $-f’’(x)/2$ $\endgroup$ – Chris Christopherson Nov 22 '19 at 10:35
  • $\begingroup$ @Luuuuuke Sorry, just noticed that myself. fixed. $\endgroup$ – user694818 Nov 22 '19 at 10:36

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