Lesson 29 (11/06/26) L'Hopital's Rule (4.7)

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Lesson 29 (11/06/26) L'Hopital's Rule (4.7)

Warmup examples:

\[ \lim_{x \to 2} \frac{x^2 - 2x}{x - 2} \quad \text{plugin?} \quad \frac{0}{0} \implies \lim_{x \to 2} \frac{x(x - 2)}{(x - 2)} = \lim_{x \to 2} x = 2 \]

\[ \lim_{x \to \infty} \frac{x^2 - 2x}{x - 2} \quad \text{plugin?} \quad \frac{\infty}{\infty} \implies \lim_{x \to \infty} \frac{x(x - 2)}{x - 2} = \lim_{x \to \infty} x = \infty \]

\[ \frac{0}{0}, \frac{\infty}{\infty} \quad \text{are called indeterminate form} \]

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L'Hopitals Rule

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

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

Example: \( \lim_{x \to 2} \frac{x^2 - 2x}{x - 2} \overset{L'H}{=} \lim_{x \to 2} \frac{\frac{d}{dx}(x^2 - 2x)}{\frac{d}{dx}(x - 2)} = \lim_{x \to 2} \frac{2x - 2}{1} = 2 \)

Example: \( \lim_{x \to \infty} \frac{x^2 - 2x}{x - 2} \overset{L'H}{=} \lim_{x \to \infty} \frac{\frac{d}{dx}(x^2 - 2x)}{\frac{d}{dx}(x - 2)} = \lim_{x \to \infty} \frac{2x - 2}{1} = \infty \)

Example: \( \lim_{x \to 2} \frac{x^2 - 2x}{(x - 2)^2} \overset{L'H}{=} \lim_{x \to 2} \frac{2x - 2}{2(x - 2)} = \frac{2}{0} \)

Note: CANNOT Apply L'H Rule

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Example:

\[ \lim_{x \to 0} \frac{1-x}{\cos x} \]

Note: Plug in \( x = 0 \): \[ \frac{1-x}{\cos x} = \frac{1-0}{\cos 0} = \frac{1}{1} = 1 \]

\[ \lim_{x \to 0} \frac{1-x}{\cos x} = 1 \]

Instructor Remark: CANNOT Apply L'Hôpital's Rule (LH)

\[ \frac{\frac{d}{dx}(1-x)}{\frac{d}{dx}(\cos x)} = \frac{-1}{\sin x} \]

Margin Note: LARGE as \( x \to 0 \)

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Example: \(\lim_{x \to \infty} \frac{\ln x}{x-1}\)

A sketch showing a coordinate plane with a vertical axis and a horizontal axis. A curve representing ln(x) is drawn, increasing slowly. The label 'ln x' is written to the left of the diagram.
Visual Description: A sketch showing a coordinate plane with a vertical axis and a horizontal axis. A curve representing ln(x) is drawn, increasing slowly. The label 'ln x' is written to the left of the diagram.

As \(x \to \infty\), \(\ln x \to \infty\) and \(x \to \infty\). This indicates the form \(\frac{\infty}{\infty}\).

Instructor Remark: CAN Apply L'H (L'Hôpital's Rule).

\[ \lim_{x \to \infty} \frac{\ln x}{x-1} = \lim_{x \to \infty} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x-1)} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1} = 0 \]

Example: \(\lim_{x \to \infty} \frac{\ln x}{5x+3} \xrightarrow{L'H} \lim_{x \to \infty} \frac{\frac{1}{x}}{5} = 0\)

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A Cartesian coordinate system plot showing two curves. One curve is a straight line labeled 'y = mx + b' with a positive slope, passing through the y-axis above the origin. The second curve is a logarithmic function labeled 'y = ln(x)', which approaches negative infinity as x approaches zero from the right and increases at a decreasing rate as x increases. The plot illustrates that for large values of x, the linear function stays above the logarithmic function.
Visual Description: A Cartesian coordinate system plot showing two curves. One curve is a straight line labeled 'y = mx + b' with a positive slope, passing through the y-axis above the origin. The second curve is a logarithmic function labeled 'y = ln(x)', which approaches negative infinity as x approaches zero from the right and increases at a decreasing rate as x increases. The plot illustrates that for large values of x, the linear function stays above the logarithmic function.

Any linear function \( mx + b \) grows much faster than \( \ln x \) as \( x \to \infty \).

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Example: L'Hôpital's Rule Application

Example: \( \lim_{x \to \infty} \frac{\ln x}{5x^3 - 7x + 3} \)

Note: The form is \( \frac{\infty}{\infty} \).

Applying L'Hôpital's Rule (denoted as L'H):

\[ \lim_{x \to \infty} \frac{\frac{1}{x}}{15x^2 - 7} = \lim_{x \to \infty} \frac{1}{x(15x^2 - 7)} = 0 \]

Observation: Any polynomial grows much faster than a log function.

\[ \lim_{x \to \infty} \frac{\ln x}{\text{polynomial}} = 0 \]

\[ \lim_{x \to \infty} \frac{\text{polynomial}}{\ln x} = \pm \infty \]

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Example:

\[ \lim_{x \to \infty} \frac{x^2}{e^x} \quad \left( \frac{\infty}{\infty} \right) \]

Applying L'Hôpital's Rule (L'H):

\[ = \lim_{x \to \infty} \frac{2x}{e^x} \quad \left( \frac{\infty}{\infty} \right) \]

Applying L'Hôpital's Rule again:

\[ \stackrel{L'H}{=} \lim_{x \to \infty} \frac{2}{e^x} = \frac{2}{\text{LARGE}} = 0 \]

Note: \( e^x \) is much larger than \( x^2 \) when \( x \) is large.

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Example:

\[ \lim_{x \to \infty} \frac{x^5 - 12x^2 + 25}{3^x} \quad \left( \frac{\infty}{\infty} \right) \]

Applying L'Hôpital's Rule (denoted as L'H):

\[ \overset{L'H}{=} \lim_{x \to \infty} \frac{5x^4 - 24x}{3^x \cdot \ln 3} \quad \left( \frac{\infty}{\infty} \right) \]

\[ \overset{L'H}{=} \lim_{x \to \infty} \frac{20x^3 - 24}{3^x \cdot (\ln 3)^2} \quad \left( \frac{\infty}{\infty} \right) \]

\[ \overset{L'H}{=} \lim_{x \to \infty} \frac{60x^2}{3^x (\ln 3)^3} \quad \left( \frac{\infty}{\infty} \right) \]

\[ \overset{L'H}{=} \lim_{x \to \infty} \frac{120x}{3^x (\ln 3)^4} \quad \left( \frac{\infty}{\infty} \right) \]

\[ \overset{L'H}{=} \lim_{x \to \infty} \frac{120}{3^x (\ln 3)^5} = \frac{120}{\text{LARGE}} = 0 \]

Reference Formulas:

\[ \frac{d}{dx} b^x = b^x \cdot \ln b \]

\[ \frac{d}{dx} 3^x = 3^x \cdot \ln 3 \]

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Logarithmic vs polynomial vs exponential

A coordinate plane showing the relative growth rates of three functions. A purple line representing an exponential function \(b^x\) is the highest curve, showing rapid growth. A green line representing a polynomial function shows moderate growth. A blue line representing a logarithmic function \(\log_b x\) is the lowest curve, showing the slowest growth rate. The x and y axes intersect in the center of the graph.
Visual Description: A coordinate plane showing the relative growth rates of three functions. A purple line representing an exponential function \(b^x\) is the highest curve, showing rapid growth. A green line representing a polynomial function shows moderate growth. A blue line representing a logarithmic function \(\log_b x\) is the lowest curve, showing the slowest growth rate. The x and y axes intersect in the center of the graph.

Exponential grows faster than any polynomial

Any polynomial grows faster than logarithmic function

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\[ \lim_{x \to 0^+} x \cdot \frac{1}{x} = 1 \]

Note: \( 0 \cdot \infty \)

\[ \lim_{x \to 0^+} x^2 \cdot \frac{1}{x} = 0 \]

Note: \( 0 \cdot \infty \)

\[ \lim_{x \to 0^+} x \cdot \frac{1}{x^2} = \infty \]

Note: \( 0 \cdot \infty \)

Other indeterminate forms

The indeterminate forms are \( 0 \cdot \infty \), \( \infty - \infty \), and \( 1^\infty \).

Instruction: Convert to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) and Apply L'H.

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Example: \(\lim_{x \to 0^+} x \cdot \ln x\)

Here, the first term \(x\) approaches \(0\), and the second term \(\ln x\) approaches \(-\infty\).

Using the substitution \(x = \frac{1}{(1/x)}\), we note that when \(x\) is small, \(\frac{1}{x}\) is large. Thus, we can write \(small = \frac{1}{(1/large)}\).

A Cartesian coordinate system plot showing a curve passing through the fourth quadrant and moving into the first quadrant. The curve represents f(x) = x ln(x), starting near the origin from the y-axis, dipping into negative values as x increases slightly, and then rising to cross the x-axis and continue upwards into the first quadrant.
Visual Description: A Cartesian coordinate system plot showing a curve passing through the fourth quadrant and moving into the first quadrant. The curve represents f(x) = x ln(x), starting near the origin from the y-axis, dipping into negative values as x increases slightly, and then rising to cross the x-axis and continue upwards into the first quadrant.

We can now evaluate the limit:

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

This results in the indeterminate form \(\frac{-\infty}{\infty}\), so we can apply L'Hôpital's Rule (L'H):

\[ = \lim_{x \to 0^+} \frac{(1/x)}{(-1/x^2)} = \lim_{x \to 0^+} \frac{-x^2}{x} = \lim_{x \to 0^+} -x = 0 \]

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Example: Evaluate the limit

Consider the limit: \[ \lim_{x \to 0^+} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] \]

Note: We observe the indeterminate form \( \infty - \infty \). To resolve this, we make it one fraction.

\[ \frac{1}{x} - \frac{1}{e^x - 1} = \frac{e^x - 1 - x}{x(e^x - 1)} \]

Note: Evaluating as \( x \to 0^+ \): \[ \frac{e^0 - 1 - 0}{0(e^0 - 1)} = \frac{0}{0} \]

Now, we return to the limit expression:

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

Applying L'Hôpital's rule (denoted by L'H):

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

Applying L'Hôpital's rule once more:

\[ \xrightarrow{L'H} \lim_{x \to 0^+} \frac{e^x}{e^x + e^x + x e^x} = \frac{1}{1 + 1 + 0} = \frac{1}{2} \]

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Example: \( \lim_{x \to \infty} (1 + \frac{1}{x})^x \)

Note: The blue ink annotations indicate the components of the limit as \( x \to \infty \): \( (1 + \frac{1}{x}) \to 1 \) and \( x \to \infty \), leading to the indeterminate form \( 1^{\infty} \).

Using the logarithm property \( \ln(m^r) = r \ln(m) \), we can write:

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

As \( x \to \infty \), this takes the form \( \infty \cdot 0 \). We rewrite this as a fraction to apply L'Hôpital's rule:

Remark: The term "rewrite" (noted as "reunto.") transforms the expression into:

\[ x \ln(1 + \frac{1}{x}) = \frac{\ln(1 + \frac{1}{x})}{\frac{1}{x}} \]

Note: The final expression is in the indeterminate form \( \frac{0}{0} \).

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Let \( L = \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x \)

\[ \ln L = \lim_{x \to \infty} x \cdot \ln\left(1 + \frac{1}{x}\right) = \lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{\left(\frac{1}{x}\right)} \]

Margin Note: The expression is in the form \(\frac{0}{0}\).

\[ \xrightarrow{H} \lim_{x \to \infty} \frac{\frac{1}{1 + \frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right)}{\left(-\frac{1}{x^2}\right)} \]

\[ = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} = 1 \]

\[ \ln L = 1 \implies e^{\ln L} = e^1 \implies L = e \]