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Question 7.T.14: (L’Hôpital’s Rule) Let the functions f and g be continuous o...

(L’Hôpital’s Rule)

Let the functions f and g be continuous on a real interval I and differentiable on I\{c}, where c is a point in I. If

(i) g^{′}(x) ≠ 0  for all x ∈ I\{c},

(ii) f(c) = g(c) = 0,

(iii) \underset{x→c}{\lim }  \frac {f^{′}(x)} {g^{′}(x)} exists in \bar{\mathbb{R}} = \mathbb{R} ∪ \left\{−∞, ∞\right\},

then

\underset{x→c}{\lim }  \frac {f(x)} {g(x)} = \underset{x→c}{\lim }  \frac {f^{′}(x)} {g^{′}(x)} .

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Let (x_{n}) be any sequence in I such that x_{n} ≠ c and x_{n} → c.

According to Theorem 5.1, it suffices to prove that

\underset{x→∞}{\lim }  \frac {f(x_{n})} {g(x_{n})} = \underset{x→c}{\lim }  \frac {f^{′}(x)} {g^{′}(x)}.

Applying Cauchy’s mean value theorem, we can find a sequence (c_{n}) in I\{c} such that

1. c_{n} lies between x_{n} and c,

2. [f(x_{n}) − f(c)]g^{′}(c_{n}) = [g(x_{n}) − g(c)]f^{′}(c_{n}) for all n ∈ \mathbb{N}.

Note that condition 1 implies g^{′}(c_{n}) ≠ 0 for all n ∈ \mathbb{N}; and since f(c) = g(c) = 0, condition 2 becomes

f(x_{n})g^{′}(c_{n}) = g(x_{n})f^{′}(c_{n}).        (7.11)

The fact that g^{′}(x) ≠ 0 on I\{c} means that g is injective on the interval whose end-points are x_{n} and c, hence

g(x_{n}) ≠ g(c) = 0.

This allows us to divide equation (7.11) by g(x_{n})g^{′}(c_{n}) to obtain

\frac {f(x_{n})} {g(x_{n})} = \frac{f^{′}(c_{n})}{g^{′}(c_{n})}.

As the sequence (f^{′}(c_{n})/g^{′}(c_{n})) converges (in the extended sense) to

\underset{x→c}{\lim }  \frac {f^{′}(x)} {g^{′}(x)},

the sequence (f(x_{n})/g(x_{n})) also converges, and to the same limit.

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