L'Hospital's Rule

So far we have learnt about and . In this chapter we will understand an important category of limit problems that can be solved easily using derivatives.

I. Introduction:

There are many cases where while computing limits of quotient of two functions, of the form:

$\lim\limits_{x \rightarrow a} \dfrac{f(x)}{g(x)}$

If the functions are such that the limits of the two functions, evaluated individually, become $0$ or $\pm \infty$, that is:

$i)\lim\limits_{x \rightarrow 0} f(x) = 0$ and $\lim\limits_{x \rightarrow 0} g(x) = 0$, or

$ii)\lim\limits_{x \rightarrow 0} f(x) = \pm \infty$ and $\lim\limits_{x \rightarrow 0} g(x) = \pm \infty$

These forms, that is, $\dfrac{0}{0}$ or $\dfrac{\pm \infty}{\pm \infty}$ are called “indeterminate forms”.

We will see the proofs for both the above cases in the following sections.

II. Proof For f(a) = 0 And g(a) = 0:

Let $f$ and $g$ be functions such that $f(a) = 0$ and $g(a) = 0$ for some value $a$.

We can write $x - a = \Delta x$, therefore,

$x = a + \Delta x$

as $x \rightarrow a$, $\Delta x \rightarrow 0$

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Therefore,

$\lim\limits_{x \rightarrow a} \dfrac{f(x)}{g(x)}$

$= \lim\limits_{x \rightarrow a} \dfrac{f(x) - 0}{g(x) - 0}$

$= \lim\limits_{x \rightarrow a} \dfrac{f(x) - f(a)}{g(x) - f(a)}$

$= \lim\limits_{x \rightarrow a} \dfrac{\dfrac{f(x) - f(a)}{x - a}}{\dfrac{g(x) - f(a)}{x - a}}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{\dfrac{f(a + \Delta x) - f(a)}{\Delta x}}{\dfrac{g(a + \Delta x) - f(a)}{\Delta x}}$

$= \dfrac{\lim\limits_{\Delta x \rightarrow 0}\dfrac{f(a + \Delta x) - f(a)}{\Delta x}}{\lim\limits_{\Delta x \rightarrow 0}\dfrac{g(a + \Delta x) - f(a)}{\Delta x}}$

$= \dfrac{f'(a)}{g'(a)}$

$= \lim\limits_{x \rightarrow a} \dfrac{f'(x)}{g'(x)}$

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III. Proof For $f(a) = \pm \infty$ And $g(a) = \pm \infty$:

Let us assume that $f(a) = \infty$ and $g(a) = \infty$

Therefore, as $x \rightarrow a$, $f(x) \rightarrow \infty \Rightarrow \dfrac{1}{f(x)} \rightarrow 0$

Similarly, as $x \rightarrow a$ $\dfrac{1}{g(x)} \rightarrow 0$

Let $L = \lim\limits_{x \rightarrow a} \dfrac{f(x)}{g(x)}$

$= \lim\limits_{x \rightarrow a} \dfrac{\dfrac{1}{g(x)}}{\dfrac{1}{f(x)}}$

$= \lim\limits_{x \rightarrow a} \dfrac{[g(x)]^{-1}}{[f(x)]^{-1}}$

Since both the numerator and denominator above tends to $0$, we get the indeterminate form $\dfrac{0}{0}$, explained in the previous section, and can apply L'Hospital's rule to this case. Therefore,

$L = \lim\limits_{x \rightarrow a} \dfrac{\dfrac{d}{dx}[g(x)]^{-1}}{\dfrac{d}{dx}[f(x)]^{-1}}$

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$\Rightarrow L = \lim\limits_{x \rightarrow a} \dfrac{-[g(x)]^{-2} \cdot \dfrac{d}{dx}g(x)}{-[f(x)]^{-2} \cdot \dfrac{d}{dx}f(x)}$

$\Rightarrow L = \lim\limits_{x \rightarrow a} \dfrac{[f(x)]^{2} \cdot g'(x)}{[g(x)]^{2} \cdot f'(x)}$

$\Rightarrow L = \lim\limits_{x \rightarrow a} \dfrac{[f(x)]^{2}}{[g(x)]^{2}} \lim\limits_{x \rightarrow a} \dfrac{g'(x)}{f'(x)}$

$\Rightarrow L = \left[\lim\limits_{x \rightarrow a} \dfrac{f(x)}{g(x)}\right]^2 \lim\limits_{x \rightarrow a} \dfrac{g'(x)}{f'(x)}$

$\Rightarrow L = \left[L\right]^2 \lim\limits_{x \rightarrow a} \dfrac{g'(x)}{f'(x)}$

$\therefore L = \lim\limits_{x \rightarrow a} \dfrac{f'(x)}{g'(x)}$

IV. Other Cases Of L'Hospital's Rule:

We saw the proofs of two indeterminate forms, for which we can apply L'Hospital's rule.

There are other more complex indeterminate forms for which we can apply this rule. These are:

- $0^0$

- $1^{\infty}$