4.8 L’hôpital’s rule (Page 2/7)

Calculus volume 1 Page 2 / 7

Proof

We provide a proof of this theorem in the special case when $f, g, f^{'},$ and $g^{'}$ are all continuous over an open interval containing $a .$ In that case, since $lim_{x \to a} f (x) = 0 = lim_{x \to a} g (x)$ and $f$ and $g$ are continuous at $a,$ it follows that $f (a) = 0 = g (a) .$ Therefore,

\begin{matrix} lim_{x \to a} \frac{f (x)}{g (x)} & = lim_{x \to a} \frac{f (x) - f (a)}{g (x) - g (a)} & since f (a) = 0 = g (a) \\ = lim_{x \to a} \frac{\frac{f (x) - f (a)}{x - a}}{\frac{g (x) - g (a)}{x - a}} & algebra \\ = \frac{lim_{x \to a} \frac{f (x) - f (a)}{x - a}}{lim_{x \to a} \frac{g (x) - g (a)}{x - a}} & limit of a quotient \\ = \frac{f^{'} (a)}{g^{'} (a)} & definition of the derivative \\ = \frac{lim_{x \to a} f^{'} (x)}{lim_{x \to a} g^{'} (x)} & continuity of f^{'} and g^{'} \\ = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} . & limit of a quotient \end{matrix}

Note that L’Hôpital’s rule states we can calculate the limit of a quotient $\frac{f}{g}$ by considering the limit of the quotient of the derivatives $\frac{f^{'}}{g^{'}} .$ It is important to realize that we are not calculating the derivative of the quotient $\frac{f}{g} .$

□

Applying l’hôpital’s rule (0/0 case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

$lim_{x \to 0} \frac{1 - cos x}{x}$
$lim_{x \to 1} \frac{sin (π x)}{ln x}$
$lim_{x \to \infty} \frac{e^{1 / x} - 1}{1 / x}$
$lim_{x \to 0} \frac{sin x - x}{x^{2}}$

Since the numerator $1 - cos x \to 0$ and the denominator $x \to 0,$ we can apply L’Hôpital’s rule to evaluate this limit. We have

$\begin{matrix} lim_{x \to 0} \frac{1 - cos x}{x} & = lim_{x \to 0} \frac{\frac{d}{d x} (1 - cos x)}{\frac{d}{d x} (x)} \\ = lim_{x \to 0} \frac{sin x}{1} \\ = \frac{lim_{x \to 0} (sin x)}{lim_{x \to 0} (1)} \\ = \frac{0}{1} = 0. \end{matrix}$
As $x \to 1,$ the numerator $sin (π x) \to 0$ and the denominator $ln (x) \to 0 .$ Therefore, we can apply L’Hôpital’s rule. We obtain

$\begin{matrix} lim_{x \to 1} \frac{sin (π x)}{ln x} & = lim_{x \to 1} \frac{π cos (π x)}{1 / x} \\ = lim_{x \to 1} (π x) cos (π x) \\ = (π \cdot 1) (−1) = − π . \end{matrix}$
As $x \to \infty,$ the numerator $e^{1 / x} - 1 \to 0$ and the denominator $(\frac{1}{x}) \to 0 .$ Therefore, we can apply L’Hôpital’s rule. We obtain

$lim_{x \to \infty} \frac{e^{1 / x} - 1}{\frac{1}{x}} = lim_{x \to \infty} \frac{e^{1 / x} (\frac{−1}{x^{2}})}{(\frac{−1}{x^{2}})} = lim_{x \to \infty} e^{1 / x} = e^{0} = 1 \cdot lim_{x \to \infty} \frac{e^{1 / x} - 1}{ln x} .$
As $x \to 0,$ both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain

$lim_{x \to 0} \frac{sin x - x}{x^{2}} = lim_{x \to 0} \frac{cos x - 1}{2 x} .$

Since the numerator and denominator of this new quotient both approach zero as $x \to 0,$ we apply L’Hôpital’s rule again. In doing so, we see that

$lim_{x \to 0} \frac{cos x - 1}{2 x} = lim_{x \to 0} \frac{− sin x}{2} = 0 .$

Therefore, we conclude that

$lim_{x \to 0} \frac{sin x - x}{x^{2}} = 0 .$

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Evaluate $lim_{x \to 0} \frac{x}{tan x} .$

$1$

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We can also use L’Hôpital’s rule to evaluate limits of quotients $\frac{f (x)}{g (x)}$ in which $f (x) \to ± \infty$ and $g (x) \to ± \infty .$ Limits of this form are classified as indeterminate forms of type $\infty / \infty .$ Again, note that we are not actually dividing $\infty$ by $\infty .$ Since $\infty$ is not a real number, that is impossible; rather, $\infty / \infty .$ is used to represent a quotient of limits, each of which is $\infty$ or $− \infty .$

L’hôpital’s rule $(\infty / \infty$ Case)

Suppose $f$ and $g$ are differentiable functions over an open interval containing $a,$ except possibly at $a .$ Suppose $lim_{x \to a} f (x) = \infty$ (or $− \infty)$ and $lim_{x \to a} g (x) = \infty$ (or $− \infty) .$ Then,

lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)},

assuming the limit on the right exists or is $\infty$ or $− \infty .$ This result also holds if the limit is infinite, if $a = \infty$ or $− \infty,$ or the limit is one-sided.

Applying l’hôpital’s rule $(\infty / \infty$ Case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

$lim_{x \to \infty} \frac{3 x + 5}{2 x + 1}$
$lim_{x \to 0^{+}} \frac{ln x}{cot x}$

Since $3 x + 5$ and $2 x + 1$ are first-degree polynomials with positive leading coefficients, $lim_{x \to \infty} (3 x + 5) = \infty$ and $lim_{x \to \infty} (2 x + 1) = \infty .$ Therefore, we apply L’Hôpital’s rule and obtain

$lim_{x \to \infty} \frac{3 x + 5}{2 x + 1} = lim_{x \to \infty} \frac{3 + 5 / x}{2 x + 1} = lim_{x \to \infty} \frac{3}{2} = \frac{3}{2} .$

Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of $x$ in the denominator. In doing so, we saw that

$lim_{x \to \infty} \frac{3 x + 5}{2 x + 1} = lim_{x \to \infty} \frac{3 + 5 / x}{2 x + 1} = \frac{3}{2} .$

L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
Here, $lim_{x \to 0^{+}} ln x = − \infty$ and $lim_{x \to 0^{+}} cot x = \infty .$ Therefore, we can apply L’Hôpital’s rule and obtain

$lim_{x \to 0^{+}} \frac{ln x}{cot x} = lim_{x \to 0^{+}} \frac{1 / x}{− {csc}^{2} x} = lim_{x \to 0^{+}} \frac{1}{− x {csc}^{2} x} .$

Now as $x \to 0^{+},$ ${csc}^{2} x \to \infty .$ Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of $csc x$ to write

$lim_{x \to 0^{+}} \frac{1}{− x {csc}^{2} x} = lim_{x \to 0^{+}} \frac{{sin}^{2} x}{− x} .$

Now $lim_{x \to 0^{+}} {sin}^{2} x = 0$ and $lim_{x \to 0^{+}} x = 0,$ so we apply L’Hôpital’s rule again. We find

$lim_{x \to 0^{+}} \frac{{sin}^{2} x}{− x} = lim_{x \to 0^{+}} \frac{2 sin x cos x}{−1} = \frac{0}{−1} = 0 .$

We conclude that

$lim_{x \to 0^{+}} \frac{ln x}{cot x} = 0 .$

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Questions & Answers

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4.8 L’hôpital’s rule (Page 2/7)

Proof

Applying l’hôpital’s rule (0/0 case)

L’hôpital’s rule ( ∞ / ∞ Case)

Applying l’hôpital’s rule ( ∞ / ∞ Case)

Questions & Answers

Read also:

Get the best Calculus volume 1 course in your pocket!

L’hôpital’s rule $(\infty / \infty$ Case)

Applying l’hôpital’s rule $(\infty / \infty$ Case)