L'Hôpital's Rule

Step 1: Check the indeterminate form

As $x \to 0$, the numerator $e^x - 1 \to e^0 - 1 = 0$ and the denominator $x \to 0$, so this is a $\dfrac{0}{0}$ indeterminate form.

Step 2: Check the conditions of L'Hôpital's rule

Check that the numerator and denominator are differentiable:

• Derivative of the numerator: $(e^x - 1)' = e^x$ (differentiable for all $x$)
• Derivative of the denominator: $(x)' = 1 \neq 0$ (the condition holds)

→ L'Hôpital's rule can be applied.

Step 3: Apply L'Hôpital's rule

Differentiate the numerator and denominator separately and compute the new limit:

$$\lim_{x \to 0} \dfrac{e^x - 1}{x} = \lim_{x \to 0} \dfrac{(e^x - 1)'}{(x)'} = \lim_{x \to 0} \dfrac{e^x}{1}$$

Step 4: Compute the new limit

Since $e^x$ is continuous at $x=0$, we can substitute directly:

$$\lim_{x \to 0} \dfrac{e^x}{1} = \dfrac{e^0}{1} = \dfrac{1}{1} = 1$$

Meaning: near $x \approx 0$, the approximation $e^x - 1 \approx x$ holds (the tangent slope of the exponential function is $1$).

Step 1: Check the indeterminate form

As $x \to 0$, the numerator $1 - \cos x \to 1 - 1 = 0$ and the denominator $x^2 \to 0$, so this is the $\dfrac{0}{0}$ form.

Step 2: First application of L'Hôpital's rule

Derivatives of the numerator and denominator:

• Numerator: $(1 - \cos x)' = 0 - (-\sin x) = \sin x$
• Denominator: $(x^2)' = 2x$

New limit:

$$\lim_{x \to 0} \dfrac{\sin x}{2x}$$

Step 3: Identify the form of the new limit

As $x \to 0$, $\sin x \to 0$ and $2x \to 0$, so it is again the $\dfrac{0}{0}$ form.

→ L'Hôpital's rule can be applied once more.

Step 4: Second application of L'Hôpital's rule

Differentiate the numerator and denominator again:

• Numerator: $(\sin x)' = \cos x$
• Denominator: $(2x)' = 2$

New limit:

$$\lim_{x \to 0} \dfrac{\cos x}{2}$$

Step 5: Compute the final limit

Since $\cos x$ is continuous at $x=0$:

$$\lim_{x \to 0} \dfrac{\cos x}{2} = \dfrac{\cos 0}{2} = \dfrac{1}{2}$$

Meaning: near $x \approx 0$, the approximation $1 - \cos x \approx \dfrac{x^2}{2}$ holds (Taylor expansion).

Step 1: Check the indeterminate form

As $x \to 0$, the numerator $\sin x \to 0$ and the denominator $x \to 0$, so this is the $\dfrac{0}{0}$ form.

Step 2: Apply L'Hôpital's rule formally

Compute the derivatives of the numerator and denominator:

• Numerator: $(\sin x)' = \cos x$
• Denominator: $(x)' = 1$

New limit:

$$\lim_{x \to 0} \dfrac{\sin x}{x} = \lim_{x \to 0} \dfrac{\cos x}{1}$$

Step 3: Compute the new limit

Since $\cos x$ is continuous:

$$\lim_{x \to 0} \dfrac{\cos x}{1} = \cos 0 = 1$$

⚠️ Important warning: the danger of circular reasoning
Proving the fact that $(\sin x)' = \cos x$ actually requires this very limit $\displaystyle\lim_{x \to 0} \dfrac{\sin x}{x} = 1$.

In other words, we fall into circular reasoning: "we use L'Hôpital's rule to find this limit, and we use this limit to justify L'Hôpital's rule."

The correct method: this limit should be proved by a geometric argument (the squeeze principle, comparing areas). For details, see Chapter 9.

Use on exams: however, it is fine to use L'Hôpital's rule to "double-check" an answer you obtained another way.

Step 1: Check the indeterminate form

As $x \to \infty$, the numerator $\ln x \to \infty$ and the denominator $x \to \infty$, so this is the $\dfrac{\infty}{\infty}$ form.

Intuitively: "the logarithm tends to infinity, and $x$ also tends to infinity. Which one goes to infinity faster?" — a race.

Step 2: Check the conditions of L'Hôpital's rule

Derivatives of the numerator and denominator:

• Numerator: $(\ln x)' = \dfrac{1}{x}$ (differentiable for $x > 0$)
• Denominator: $(x)' = 1 \neq 0$ (the condition holds)

→ L'Hôpital's rule can be applied.

Step 3: Apply L'Hôpital's rule

$$\lim_{x \to \infty} \dfrac{\ln x}{x} = \lim_{x \to \infty} \dfrac{(\ln x)'}{(x)'} = \lim_{x \to \infty} \dfrac{1/x}{1} = \lim_{x \to \infty} \dfrac{1}{x}$$

Step 4: Compute the new limit

As $x \to \infty$, $\dfrac{1}{x} \to 0$, so:

$$\lim_{x \to \infty} \dfrac{1}{x} = 0$$

Meaning: the logarithm $\ln x$ tends to infinity, but the linear function $x$ tends to infinity much faster. In other words, "$x$ is overwhelmingly stronger than $\ln x$."

Step 1: Check the indeterminate form and the strategy

As $x \to \infty$, the numerator $e^x \to \infty$ and the denominator $x^n \to \infty$, so this is the $\dfrac{\infty}{\infty}$ form.

The intuitive question: "the exponential versus the polynomial — which goes to infinity faster?"

Strategy: the numerator stays $e^x$ no matter how many times we differentiate. The denominator is a polynomial, so after differentiating $n$ times it becomes a constant tending to zero in degree. Hence applying L'Hôpital's rule $n$ times settles it.

Step 2: Apply L'Hôpital's rule repeatedly

First differentiation:

$$\lim_{x \to \infty} \dfrac{e^x}{x^n} = \lim_{x \to \infty} \dfrac{(e^x)'}{(x^n)'} = \lim_{x \to \infty} \dfrac{e^x}{nx^{n-1}}$$

This is still the $\dfrac{\infty}{\infty}$ form, so we differentiate a second time:

$$= \lim_{x \to \infty} \dfrac{e^x}{n(n-1)x^{n-2}}$$

And we continue... (each time the degree of the denominator drops by one)

Step 3: The form after $n$ differentiations

After applying L'Hôpital's rule $n$ times, the polynomial in the denominator is entirely consumed and becomes the constant $n! = n(n-1)(n-2)\cdots 2 \cdot 1$:

$$\lim_{x \to \infty} \dfrac{e^x}{x^n} = \lim_{x \to \infty} \dfrac{e^x}{n!}$$

Step 4: Compute the final limit

The numerator $e^x \to \infty$ and the denominator $n!$ is a constant, so:

$$\lim_{x \to \infty} \dfrac{e^x}{n!} = \infty$$

Conclusion: the exponential $e^x$ tends to infinity faster than any power function $x^n$. This is an extremely important property in mathematics and physics.

Step 1: Check the indeterminate form

As $x \to 0^+$:

• $x \to 0^+$ (tends to zero)
• $\ln x \to -\infty$ (tends to negative infinity)

so this is the $0 \cdot (-\infty)$ form.

Step 2: Convert the indeterminate form into a fraction

To turn $x \ln x$ into a fraction, put one factor in the numerator and the reciprocal of the other in the denominator. Here we put $\ln x$ in the numerator and $1/x$ in the denominator:

$$\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \dfrac{\ln x}{1/x}$$

Step 3: Check the form of the new limit

As $x \to 0^+$:

• Numerator: $\ln x \to -\infty$
• Denominator: $\dfrac{1}{x} \to +\infty$

$\dfrac{-\infty}{+\infty}$ is formally the $\dfrac{\infty}{\infty}$ form. → L'Hôpital's rule applies!

Step 4: Apply L'Hôpital's rule

Compute the derivatives of the numerator and denominator:

• Numerator: $(\ln x)' = \dfrac{1}{x}$
• Denominator: $\left(\dfrac{1}{x}\right)' = -\dfrac{1}{x^2}$

New limit:

$$= \lim_{x \to 0^+} \dfrac{(\ln x)'}{(1/x)'} = \lim_{x \to 0^+} \dfrac{1/x}{-1/x^2}$$

Step 5: Simplify the fraction

Multiply numerator and denominator by $x^2$ and simplify:

$$\lim_{x \to 0^+} \dfrac{1/x}{-1/x^2} = \lim_{x \to 0^+} \dfrac{1}{x} \cdot \dfrac{x^2}{-1} = \lim_{x \to 0^+} \dfrac{x^2}{-x} = \lim_{x \to 0^+} (-x)$$

Step 6: Compute the final limit

As $x \to 0^+$, $-x \to 0$, so:

$$\lim_{x \to 0^+} (-x) = 0$$

Meaning: $0$ is "stronger" than $\infty$, so in the end it converges to $0$. Intuitively, when $x$ is very small, the smallness of $x$ wins over the negative magnitude of $\ln x$.

Step 1: Check the indeterminate form

As $x \to 0^+$:

• $\dfrac{1}{\sin x} \to +\infty$ (positive infinity)
• $\dfrac{1}{x} \to +\infty$ (positive infinity)

so this is the $\infty - \infty$ indeterminate form. Intuitively it is the question "which one tends to infinity faster?"

Step 2: Combine into a single fraction

To turn the subtraction into a fraction, combine over a common denominator:

$$\lim_{x \to 0^+} \left(\dfrac{1}{\sin x} - \dfrac{1}{x}\right) = \lim_{x \to 0^+} \dfrac{x - \sin x}{x \sin x}$$

Step 3: Check the form of the new limit

As $x \to 0^+$:

• Numerator: $x - \sin x \to 0 - 0 = 0$
• Denominator: $x \sin x \to 0 \cdot 0 = 0$

We have converted it to the $\dfrac{0}{0}$ form. → L'Hôpital's rule applies.

Step 4: Apply L'Hôpital's rule (first time)

Compute the derivatives of the numerator and denominator:

• Derivative of the numerator: $(x - \sin x)' = 1 - \cos x$
• Derivative of the denominator: $(x \sin x)' = \sin x + x \cos x$ (product rule)

New limit:

$$= \lim_{x \to 0^+} \dfrac{1 - \cos x}{\sin x + x \cos x}$$

Step 5: Check the form of the new limit

As $x \to 0^+$:

• Numerator: $1 - \cos x \to 1 - 1 = 0$
• Denominator: $\sin x + x \cos x \to 0 + 0 \cdot 1 = 0$

Again the $\dfrac{0}{0}$ form. → L'Hôpital's rule can be applied once more.

Step 6: Apply L'Hôpital's rule (second time)

Differentiate the numerator and denominator once more:

• Derivative of the numerator: $(1 - \cos x)' = \sin x$
• Derivative of the denominator: $(\sin x + x \cos x)' = \cos x + (\cos x - x\sin x) = 2\cos x - x\sin x$

New limit:

$$= \lim_{x \to 0^+} \dfrac{\sin x}{2\cos x - x\sin x}$$

Step 7: Compute the final limit

As $x \to 0^+$, both the numerator and the denominator tend to definite values:

• Numerator: $\sin x \to 0$
• Denominator: $2\cos x - x\sin x \to 2 \cdot 1 - 0 = 2$

Therefore:

$$\lim_{x \to 0^+} \dfrac{\sin x}{2\cos x - x\sin x} = \dfrac{0}{2} = 0$$

Meaning: near $x \approx 0$, $\dfrac{1}{\sin x}$ and $\dfrac{1}{x}$ are very close in value, but $\dfrac{1}{\sin x}$ is slightly larger, so the result of the subtraction converges to $0$.

Step 1: Check the indeterminate form

As $x \to 0^+$, $x^x$ is the $0^0$ form, where "the base tends to $0$ and the exponent also tends to $0$."

Intuitively: "what is zero to the zeroth power?"

Step 2: Take the logarithm to bring down the exponent

Set $y = x^x$ and take the natural logarithm of both sides:

$$\ln y = \ln(x^x) = x \ln x$$

Step 3: Find the limit of $\ln y$

Now it suffices to find the limit of $x \ln x$. We already computed this in Example 6:

$$\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} x \ln x = 0$$

Step 4: Find the limit of the original $y$

If $\lim_{x \to 0^+} \ln y = 0$, then the limit of the original $y$ is:

$$\lim_{x \to 0^+} y = e^{\lim_{x \to 0^+} \ln y} = e^0 = 1$$

Conclusion: $\displaystyle\lim_{x \to 0^+} x^x = 1$. That is, in the sense of limits $0^0$ is defined to be $1$.

Step 1: Check the indeterminate form

As $x \to \infty$:

• Base: $1 + \dfrac{1}{x} \to 1 + 0 = 1$ (tends to $1$)
• Exponent: $x \to \infty$ (tends to infinity)

the $1^\infty$ indeterminate form. Intuitively: "what is something that is almost exactly $1$ raised to an infinite power?"

Step 2: Take the logarithm to bring down the exponent

Set $y = \left(1 + \dfrac{1}{x}\right)^x$ and take the natural logarithm:

$$\ln y = x \ln\left(1 + \dfrac{1}{x}\right)$$

Step 3: Recognize the form of $\ln y$

As $x \to \infty$:

• $x \to \infty$ (infinity)
• $\ln\left(1 + \dfrac{1}{x}\right) \to \ln 1 = 0$ (zero)

the $\infty \cdot 0$ form. This is the same as $0 \cdot \infty$.

Step 4: Substitute to convert to the $\dfrac{0}{0}$ form

Let $t = \dfrac{1}{x}$; then $t \to 0^+$ as $x \to \infty$.

Also $x = \dfrac{1}{t}$, so:

$$\lim_{x \to \infty} x \ln\left(1 + \dfrac{1}{x}\right) = \lim_{t \to 0^+} \dfrac{1}{t} \ln(1+t) = \lim_{t \to 0^+} \dfrac{\ln(1+t)}{t}$$

This has been reduced to the $\dfrac{0}{0}$ form.

Step 5: Apply L'Hôpital's rule

Compute the derivatives of the numerator and denominator:

• Numerator: $(\ln(1+t))' = \dfrac{1}{1+t}$
• Denominator: $(t)' = 1$

New limit:

$$\lim_{t \to 0^+} \dfrac{\ln(1+t)}{t} = \lim_{t \to 0^+} \dfrac{1/(1+t)}{1} = \lim_{t \to 0^+} \dfrac{1}{1+t}$$

Step 6: Compute the final limit

As $t \to 0^+$:

$$\lim_{t \to 0^+} \dfrac{1}{1+t} = \dfrac{1}{1+0} = 1$$

Step 7: Find the limit of the original $y$

If $\lim_{x \to \infty} \ln y = 1$, then:

$$\lim_{x \to \infty} y = e^1 = e$$

Conclusion: $\displaystyle\lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x = e$. This is the definition of the base of the natural logarithm, $e \approx 2.71828\ldots$.

❌ Why L'Hôpital's rule cannot be used

At $x \to 0$ it superficially looks like the $\dfrac{0}{0}$ form, but…

Differentiating the numerator:

$$(x^2 \sin(1/x))' = 2x \sin(1/x) - \cos(1/x)$$

Derivative of the denominator: $(x)' = 1$

As $x \to 0$, $2x\sin(1/x) \to 0$, but $\cos(1/x)$ oscillates forever between $-1$ and $1$, so $\displaystyle\lim \dfrac{f'(x)}{g'(x)}$ does not exist.

⟹ The conditions of L'Hôpital's rule are not satisfied, so it cannot be used!

✓ Correct method: use a different approach

Use the squeeze principle. Since $\left|\sin(1/x)\right| \leq 1$:

$$\left|\dfrac{x^2 \sin(1/x)}{x}\right| = |x \sin(1/x)| \leq |x| \to 0 \quad (x \to 0)$$

Therefore $\displaystyle\lim_{x \to 0} \dfrac{x^2 \sin(1/x)}{x} = 0$