Paper 1, Section II, D

Analysis I | Part IA, 2009

State and prove Rolle's theorem.

Let ff and gg be two continuous, real-valued functions on a closed, bounded interval [a,b][a, b] that are differentiable on the open interval (a,b)(a, b). By considering the determinant

ϕ(x)=∣110f(a)f(b)f(x)g(a)g(b)g(x)∣=g(x)(f(b)−f(a))−f(x)(g(b)−g(a))\phi(x)=\left|\begin{array}{ccc} 1 & 1 & 0 \\ f(a) & f(b) & f(x) \\ g(a) & g(b) & g(x) \end{array}\right|=g(x)(f(b)-f(a))-f(x)(g(b)-g(a))

or otherwise, show that there is a point c∈(a,b)c \in(a, b) with

f′(c)(g(b)−g(a))=g′(c)(f(b)−f(a))f^{\prime}(c)(g(b)-g(a))=g^{\prime}(c)(f(b)-f(a))

Suppose that f,g:(0,∞)→Rf, g:(0, \infty) \rightarrow \mathbb{R} are differentiable functions with f(x)→0f(x) \rightarrow 0 and g(x)→0g(x) \rightarrow 0 as x→0x \rightarrow 0. Prove carefully that if the limit⁡x→0f′(x)g′(x)=ℓ\operatorname{limit}_{x \rightarrow 0} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\ell exists and is finite, then the limit lim⁡x→0f(x)g(x)\lim _{x \rightarrow 0} \frac{f(x)}{g(x)} also exists and equals ℓ\ell.

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