If $\theta(h)$ is such that $f(3+h)-f(3) = hf'(3+\theta(h)h)$, what is $\lim_{h \to 0} θ(h)$?

Question

Problem. Let $f(x) = \sqrt[3]{x}$ for $x\in(0,\infty)$, and $\theta(h)$ be a function such that $$f(3+h)-f(3) = hf'(3+\theta(h)h)$$ for all $h \in (−1, 1)$. Then $\lim_{h \to 0} θ(h)$ is equal to? (rounded off to two decimal places).

This question is from IIT JAM exam.

First Approach : I tried it to equate with the expression of Taylor's series for $f$ at $x=3$ but cannot conclude anything . Second Approach: Lagrange's Mean Value theorem , rearranged this expression in the form of $f'$ then evaluated derivative of $f$ at $3$ but there is no term containing $h$ & that is different from the actual answer. Both these approaches are almost same & I cannot proceed further evaluating limit. Any suggestion how to proceed ?

Answer 1

Problem. Let $\,f(x)=3\sqrt x\,$ for $\,x\in(0,\infty)\,$ and $\,\theta(h)$ be a function such that $$f(3+h)-f(3)=hf'\big(3+\theta(h)h\big)$$ for all $\,h\in(−1,1)\,.\;$ Then $\,\lim\limits_{h\to0}\theta(h)\,$ is equal to ? (rounded off to two decimal places).

Solution:

Since $\;f(x)=3\sqrt x\;$ for any $\;x\in(0,+\infty)\;$ and
$f(3+h)-f(3)=hf'\big(3+\theta(h)h\big)\;$ for all $\,h\in(−1, 1)\;,$
it follows that

$3\sqrt{3+h}-3\sqrt3=\dfrac{3h}{2\sqrt{3+\theta(h)h}}\;\;$ for all $\;h\in(-1,1)\;\;,$

$3\left(\sqrt{3+h}-\sqrt3\right)\left(\sqrt{3+h}+\sqrt3\right)=\dfrac{3h\left(\sqrt{3+h}+\sqrt3\right)}{2\sqrt{3+\theta(h)h}}\;\;,$

$3h=\dfrac{3h\left(\sqrt{3+h}+\sqrt3\right)}{2\sqrt{3+\theta(h)h}}\;\;$ for all $\;h\in(-1,1)\;\;,$

$\sqrt{3+\theta(h)h}=\dfrac{\sqrt{3+h}+\sqrt3}2\;\;$ for all $\;h\in(-1,1)\setminus\{0\}\;\;,$

$\theta(h)h=\left(\!\dfrac{\sqrt{3+h}+\sqrt3}2\right)^{\!2}\!-3\;\;$ for all $\;h\in(-1,1)\setminus\{0\}\;\;,$

$\theta(h)h=\left(\!\dfrac{\sqrt{3+h}+\sqrt3}2+\sqrt3\!\right)\!\left(\!\dfrac{\sqrt{3+h}+\sqrt3}2-\sqrt3\!\right)\;\;,$

$\theta(h)h=\left(\!\dfrac{\sqrt{3+h}+3\sqrt3}2\!\right)\!\left(\!\dfrac{\sqrt{3+h}-\sqrt3}2\!\right)\;\;,$

$\theta(h)=\left(\!\dfrac{\sqrt{3+h}+3\sqrt3}2\!\right)\!\left(\!\dfrac{\sqrt{3+h}-\sqrt3}{2h}\!\right)\;\;,$

$\theta(h)=\left(\!\dfrac{\sqrt{3+h}+3\sqrt3}2\!\right)\!\left[\dfrac1{2\left(\sqrt{3+h}+\sqrt3\right)}\right]\;\;,$

$\theta(h)=\dfrac{\sqrt{3+h}+3\sqrt3}{4\left(\sqrt{3+h}+\sqrt3\right)}\;\;$ for all $\;h\in(-1,1)\setminus\{0\}\;.$

Consequently,

$\lim\limits_{h\to0}\theta(h)=\dfrac{4\sqrt3}{8\sqrt3}=\dfrac12\;.$

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We can generalize the problem, but it is not necessary that the function $\,f\,$ is twice continuously differentiable. It is sufficient that it is twice differentiable just at one point.

Generalization of the problem:

Let $\,a,b,c,r\in\Bbb R\,$ such that $\,a<b\,,\,r>0\,$ and $\,(c-r,c+r)\subseteq(a,b)\,.\,$ Let $\,f:(a,b)\to\Bbb R\,$ be a differentiable function on $\,(c-r,c+r)\,.$
If there exists $\,f’’(c)\neq0\,$ and if $\,\theta:(-r,r)\to(0,1)\,$ is a function such that $$f(c+h)-f(c)=hf'\big(c+\theta(h)h\big)$$ for all $\,h\in(−r,r)\,.\;$ Then $\,\lim\limits_{h\to0}\theta(h)\,$ is equal to ? (rounded off to two decimal places).

Solution:

For any $\,h\in(-r,r)\setminus\{0\}\,,\,$ it results that

$h\big[f'\big(c+\theta(h)h\big)-f’(c)\big]=f(c+h)-f(c)-hf’(c)\;\;,$

$\theta(h)\!\left[\dfrac{f'\!\big(c\!+\!\theta(h)h\big)\!-\!f’(c)}{\theta(h)h}\right]=\dfrac{f(c\!+\!h)\!-\!f(c)\!-\!hf’(c)}{h^2}\;\;.$

By applying L'Hôpital's rule and by using the definition of second derivative at a point, we get that

$\lim\limits_{h\to0}\dfrac{f(c\!+\!h)\!-\!f(c)\!-\!hf’\!(c)}{h^2}=\lim\limits_{h\to0}\dfrac{f’\!(c\!+\!h)\!-\!f’\!(c)}{2h}=\dfrac{f’’\!(c)}2\,.$

Moreover, by definition of second derivative, we get that

$\lim\limits_{h\to0}\dfrac{f'\big(c+\theta(h)h\big)-f’(c)}{\theta(h)h}=f’’(c)\;.$

Since $\,f’’(c)\neq0\,,\,$ it follows that

$\lim\limits_{h\to0}\theta(h)=\lim\limits_{h\to0}\dfrac{\frac{f(c+h)-f(c)-hf’(c)}{h^2}}{\frac{f'\big(c+\theta(h)h\big)-f’(c)}{\theta(h)h}}=\dfrac{\frac{f’’(c)}2}{f’’(c)}=\dfrac12\;.$

Answer 2

We consider a slight generalization of the problem, since this clarifies the structure of the problem without introducing too much abstraction.

Let $f$ be twice continuously differentiable on an interval $(a-r, a+r)$ with $f''(a) \neq 0$, and let $\theta(h)$ be defined for $|h| < r$ by the relation

$$ f(a+h) - f(a) = h f'(a + \theta(h)h). $$

By Taylor's approximation, there exists $\xi_0 = \xi_0(h)$ and $\xi_1 = \xi_1(h)$ such that $|\xi_i(h)| \leq |h|$ and the following equalities hold:

$$ f(a + h) = f(a) + f'(a)h + f''(a+\xi_0(h))\frac{h^2}{2!}, $$

$$ f'(a + h) = f'(a) + f''(a + \xi_1(h))h. $$

Plugging this to the defining equation of $\theta(h)$, it follows that

$$ f'(a)h + f''(a+\xi_0(h))\frac{h^2}{2!} = h \left[ f'(a) + f''(a + \xi_1(\theta(h) h)) \theta(h) h \right]. $$

Rearranging this for $\theta(h)$ and taking limit as $h \to 0$,

$$ \theta(h) = \frac{1}{2} \frac{f''(a+\xi_0(h))}{f''(a + \xi_1(\theta(h) h))} \to \frac{1}{2} \frac{f''(a)}{f''(a)} = \frac{1}{2} $$

since $\xi_i(h) \to 0$ as $h \to 0$ and $f''$ is continuous at $a$ with non-zero value.