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4.7 l'Hospital's Rule

revisit an old problem: \(\lim_{x \to 2} \frac{x^2+x-6}{x^2-4} \to \frac{4+2-6}{4-4} \to \frac{0}{0}\) indeterminate form of \(\frac{0}{0}\)

old way: factoring and canceling

\[\lim_{x \to 2} \frac{(x+3)(x-2)}{(x+2)(x-2)} = \lim_{x \to 2} \frac{x+3}{x+2} = \frac{5}{4}\]

today, we'll learn l'Hospital's Rule to handle \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) when computing limits

\(\frac{0}{0}\)

\(\frac{\text{small #}}{\text{small #}}\)

\(\frac{\infty}{\infty}\)

\(\frac{\text{big #}}{\text{big #}}\)

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l'Hospital's Rule

if \(\lim_{x \to a} \frac{f(x)}{g(x)} \to \frac{0}{0}\) or \(\frac{\infty}{\infty}\)

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

named after Marquis de l'Hospital
(old spelling l'Hôpital)
1661-1704
published the first calculus textbook in 1696

example

\(\lim_{x \to 2} \frac{x^2+x-6}{x^2-4} \to \frac{0}{0}\) so we can use l'Hospital's Rule

this L means I used l'Hospital's Rule this step

\(\stackrel{L}{=} \lim_{x \to 2} \frac{2x+1}{2x}\)

now let \(x=2\)

\[= \frac{2(2)+1}{2(2)} = \frac{5}{4}\]

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L'Hôpital's Rule Examples

Example 1

\[ \lim_{x \to 0} \frac{x + \sin 2x}{x - \sin 2x} \xrightarrow{\text{as } x \to 0} \frac{0}{0} \]

MUST see \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) to use L'Hôpital's Rule

\[ \stackrel{L}{=} \lim_{x \to 0} \frac{1 + \cos(2x) \cdot 2}{1 - \cos(2x) \cdot 2} \]

now try letting \( x = 0 \)

\[ = \frac{1 + \cos(0) \cdot 2}{1 - \cos(0) \cdot 2} = \frac{1 + 2}{1 - 2} = \frac{3}{-1} = \boxed{-3} \]

Example 2

\[ \lim_{x \to 0} \frac{x + \sin 2x}{x - \cos 2x} \xrightarrow{\text{as } x \to 0} \frac{0}{0 - 1} = \frac{0}{-1} \]

NOT \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)

DO NOT use L'Hôpital's Rule

\[ = \frac{0}{-1} = \boxed{0} \]

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Growth Rates of Functions

Example

Which of the following grows faster as \( x \to \infty \)?

\[ f(x) = x^2 \quad \text{or} \quad g(x) = 2^x \, ? \]

\[ \text{if } \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{x^2}{2^x} = 0 \]

then that means \( 2^x \) eventually becomes much larger than \( x^2 \) so that also means \( 2^x \) grows faster.

\[ \text{if } \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{x^2}{2^x} = \infty \]

then \( x^2 \) eventually is much bigger so must be growing faster

\[ \text{if } \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{x^2}{2^x} \to \frac{\infty}{\infty} = \, ? \]

we can use L'Hôpital's Rule to see what is going on

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L'Hospital's Rule Example

\[ \lim_{x \to \infty} \frac{x^2}{2^x} \xrightarrow{\text{as } x \to \infty} \frac{\infty}{\infty} = ? \]

Since it is \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \) we can use L'Hospital's Rule.

\[ \stackrel{L}{=} \lim_{x \to \infty} \frac{2x}{2^x \cdot \ln 2} \xrightarrow{\text{as } x \to \infty} \frac{\infty}{\infty} \]

\[ \frac{d}{dx}(a^x) = a^x \cdot \ln a \]

After applying L'Hospital's Rule, the limit remains \( \frac{\infty}{\infty} \) (or \( \frac{0}{0} \)). When this happens, use the Rule again (until the limit is no longer \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \)).

\[ \stackrel{L}{=} \lim_{x \to \infty} \frac{2}{(2^x \cdot \ln 2) \cdot \ln 2} \xrightarrow{\text{as } x \to \infty} \frac{2}{\infty} \]

deriv. of \( 2^x \)

Do NOT use L'Hospital's Rule again since it is not \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \).

\[ = 0 \]

so \( 2^x \) grows faster than \( x^2 \).

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Other Indeterminate Forms

\( \infty - \infty \) (not necessarily 0; means big # - big #)
\( 1^\infty \) ((# close to 1)\(^{\text{big #}}\))
\( \infty \cdot 0 \) (means (big #)(small #))

Example

\[ \lim_{x \to 0^+} \left( \frac{1}{x} - \frac{1}{e^x - 1} \right) \to \infty - \infty = ? \]

Transform into \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) then use L'Hospital.

\[ = \lim_{x \to 0^+} \left( \frac{(e^x - 1)}{x \cdot (e^x - 1)} - \frac{x}{x \cdot (e^x - 1)} \right) \]\[ = \lim_{x \to 0^+} \frac{e^x - 1 - x}{x(e^x - 1)} \xrightarrow{\text{as } x \to 0} \frac{0}{0} \]

Now we can use L'Hospital.

\[ \stackrel{L}{=} \lim_{x \to 0^+} \frac{e^x - 1}{x e^x + (e^x - 1)} \xrightarrow{\text{as } x \to 0} \frac{0}{0} \]

L'Hospital again

\[ \stackrel{L}{=} \lim_{x \to 0^+} \frac{e^x}{x e^x + e^x + e^x} \xrightarrow{\text{as } x \to 0} \frac{1}{2} \]

This is the limit.

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Example: Indeterminate Forms

\[ \lim_{x \to 0^+} (1+x)^{\cot x} \xrightarrow{\text{as } x \to 0} 1^{\infty} = ? \]

Transform into \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) then use L'Hospital's Rule.

\[ \lim_{x \to 0^+} \underbrace{(1+x)^{\cot x}}_{y} \]

So, we want to know \( \lim_{x \to 0^+} y \).

Logarithmic Transformation

\[ \ln y = \ln (1+x)^{\cot x} \]\[ = \cot x \cdot \ln(1+x) = \frac{\ln(1+x)}{\tan x} \]

Now notice as \( x \to 0^+ \):

\[ \lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} \frac{\ln(1+x)}{\tan x} \to \frac{0}{0} \]

Applying L'Hospital's Rule

\[ \lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} \frac{\ln(1+x)}{\tan x} \stackrel{L}{=} \lim_{x \to 0^+} \frac{\frac{1}{1+x}}{\sec^2 x} \xrightarrow{\text{as } x \to 0} \frac{1}{1} = 1 \]

Not done yet!

We want \( \lim_{x \to 0^+} y \) but found \( \lim_{x \to 0^+} \ln y = 1 \).

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Final Result

We know \( y = e^{\ln y} \).

So, if \( \ln y \to 1 \), then \( y \to e^1 \to e \).

So, the limit is e

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Example: Evaluating Limits with L'Hôpital's Rule

Evaluate the following limit:

\[ \lim_{x \to \infty} x^2 \left( \frac{1}{x} - \sin \frac{1}{x} \right) \]

As \( x \to \infty \), the expression takes the indeterminate form \( \infty \cdot 0 = ? \)

Strategy: Transform into \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) then use L'Hôpital's Rule.

\[ = \lim_{x \to \infty} x^2 \left( \frac{1}{x} - \sin \frac{1}{x} \right) \]

Note:

\[ x^2 = \frac{1}{\frac{1}{x^2}} \]

\[ = \lim_{x \to \infty} \frac{\frac{1}{x} - \sin \frac{1}{x}}{\frac{1}{x^2}} \]

As \( x \to \infty \), this results in the form \( \frac{0}{0} \). Now use L'Hôpital's Rule.

(Some steps of derivation skipped)

\[ \stackrel{L}{=} \lim_{x \to \infty} \frac{1 - \cos(\frac{1}{x})}{\frac{2}{x}} \]

As \( x \to \infty \), this still results in the form \( \frac{0}{0} \). Apply L'Hôpital's Rule again.

(Derivation steps skipped)

\[ \stackrel{L}{=} \lim_{x \to \infty} \frac{\sin(\frac{1}{x})}{2} \]

As \( x \to \infty \), this results in \( \frac{0}{2} \). No more L'Hôpital's Rule needed.

= 0

That's the limit.