Examples of L'Hôpital's Rule: 3 Practical Examples (Plus More)

Starting with the core idea (through examples)

Before we even say the rule out loud, let’s look at a classic example of L’Hôpital’s Rule in action:

You’re asked to find
[
\lim_{x \to 0} \frac{\sin x}{x}.
]

If you plug in \(x = 0\), you get \(0/0\). That’s an indeterminate form: it doesn’t tell you whether the limit should be 0, 1, or blow up to infinity. This is exactly the kind of situation where L’Hôpital’s Rule shines.

L’Hôpital says (informally): if your limit is of the form 0/0 or ∞/∞, you’re allowed to take the derivative of the top and the derivative of the bottom separately, and try the limit again.

For this first example of L’Hôpital’s Rule:

• Numerator: \(\sin x\) → derivative is \(\cos x\).
• Denominator: \(x\) → derivative is 1.

So we transform the original limit into
[
\lim_{x \to 0} \frac{\sin x}{x} \overset{\text{L’Hôpital}}{=} \lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1.
]

That’s our first of many examples of L’Hôpital’s Rule: a clean, single‑step move from 0/0 to a nice, finite answer.

Example 1: A classic 0/0 limit you’ll see on every test

Let’s warm up with something that shows up in nearly every calculus course:
[
\lim_{x \to 1} \frac{x^2 - 1}{x - 1}.
]

Plug in \(x = 1\):

• Numerator: \(1^2 - 1 = 0\)
• Denominator: \(1 - 1 = 0\)

Again, we have 0/0. One way is to factor \(x^2 - 1 = (x-1)(x+1)\), cancel, and get the limit of \(x+1\), which is 2. But we’re here for L’Hôpital’s Rule, so let’s treat this as another example of L’Hôpital’s Rule: 3 practical examples will all build from this style.

Differentiate top and bottom:

• \(\frac{d}{dx}(x^2 - 1) = 2x\)
• \(\frac{d}{dx}(x - 1) = 1\)

Apply the rule:
[
\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{2x}{1} = 2.
]

Same answer, different tool. This is one of the best examples to see how L’Hôpital’s Rule turns an algebra headache into a quick derivative.

Example 2: Exponential vs. linear growth (and a real‑world flavor)

Now let’s look at a limit that feels more like something you might see in a model of population growth or interest rates:
[
\lim_{x \to 0} \frac{e^x - 1}{x}.
]

Again, plug in \(x = 0\):

• Numerator: \(e^0 - 1 = 1 - 1 = 0\)
• Denominator: \(0\)

We get 0/0, so it’s a candidate for L’Hôpital’s Rule.

Differentiate numerator and denominator:

• \(\frac{d}{dx}(e^x - 1) = e^x\)
• \(\frac{d}{dx}(x) = 1\)

Apply L’Hôpital:
[
\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1.
]

Why does this matter? In calculus and differential equations, this limit shows up whenever you linearize an exponential process near time zero: for very small times, the growth rate of \(e^x\) is approximately 1 per unit of \(x\). You’ll see this in models of interest growth, population dynamics, and even in some pharmacokinetics models discussed in resources like the National Library of Medicine.

This is another core example of L’Hôpital’s Rule: 3 practical examples like this one will cover most of the standard limits you meet early on.

Example 3: A trigonometric limit with a twist

Try this one:
[
\lim_{x \to 0} \frac{1 - \cos x}{x^2}.
]

Plug in \(x = 0\):

• Numerator: \(1 - \cos 0 = 1 - 1 = 0\)
• Denominator: \(0^2 = 0\)

We’re back to 0/0. Perfect candidate.

Differentiate numerator and denominator once:

• Numerator: \(\frac{d}{dx}(1 - \cos x) = \sin x\)
• Denominator: \(\frac{d}{dx}(x^2) = 2x\)

Now consider
[
\lim_{x \to 0} \frac{\sin x}{2x}.
]

Plug in \(x = 0\) again: still 0/0. So we apply L’Hôpital’s Rule a second time:

• \(\frac{d}{dx}(\sin x) = \cos x\)
• \(\frac{d}{dx}(2x) = 2\)

Now the limit becomes
[
\lim_{x \to 0} \frac{\cos x}{2} = \frac{\cos 0}{2} = \frac{1}{2}.
]

This is a nice example of L’Hôpital’s Rule where you need more than one round. The pattern is simple: if you still get 0/0 or ∞/∞ after one application, you’re allowed to repeat as long as the conditions hold.

Going beyond 3: more real examples of L’Hôpital’s Rule in action

So far we’ve walked through three core examples of L’Hôpital’s Rule: 3 practical examples that cover algebraic, exponential, and trigonometric cases. Now let’s expand the toolbox with more situations you’ll actually run into.

Logarithmic growth vs. linear growth

Consider
[
\lim_{x \to \infty} \frac{\ln x}{x}.
]

As \(x \to \infty\):

• \(\ln x \to \infty\)
• \(x \to \infty\)

So this is an ∞/∞ form. Differentiate top and bottom:

• \(\frac{d}{dx}(\ln x) = \frac{1}{x}\)
• \(\frac{d}{dx}(x) = 1\)

Apply L’Hôpital:
[
\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0.
]

Interpretation: linear growth beats logarithmic growth in the long run. This kind of comparison shows up in computer science (analyzing algorithms) and data science when you’re comparing model complexities.

Exponential dominates polynomial

Now look at
[
\lim_{x \to \infty} \frac{x^3}{e^x}.
]

Again, both numerator and denominator go to infinity, so we have ∞/∞. Apply L’Hôpital’s Rule repeatedly:

First round:

• Top: \(3x^2\)
• Bottom: \(e^x\)

Second round:

• Top: \(6x\)
• Bottom: \(e^x\)

Third round:

• Top: 6
• Bottom: \(e^x\)

Now
[
\lim_{x \to \infty} \frac{6}{e^x} = 0.
]

You’ll see this pattern a lot: exponential terms like \(e^x\) eventually crush any polynomial. This is one of the best examples of L’Hôpital’s Rule for comparing long‑term growth rates.

Ratio of two cooling processes (physics / engineering flavor)

Suppose two objects are cooling according to simplified models:

• Object A: temperature difference from room temperature behaves like \(T_A(t) = e^{-t}\)
• Object B: \(T_B(t) = e^{-2t}\)

You might want the limit of their ratio as time goes on:
[
\lim_{t \to \infty} \frac{T_A(t)}{T_B(t)} = \lim_{t \to \infty} \frac{e^{-t}}{e^{-2t}} = \lim_{t \to \infty} e^{t}.
]

This one actually blows up to \(\infty\) without any need for L’Hôpital. But let’s tweak it into a form where L’Hôpital’s Rule does apply:
[
\lim_{t \to 0} \frac{e^{-t} - 1}{e^{-2t} - 1}.
]

As \(t \to 0\), both numerator and denominator go to 0. Differentiate:

• Numerator: \(\frac{d}{dt}(e^{-t} - 1) = -e^{-t}\)
• Denominator: \(\frac{d}{dt}(e^{-2t} - 1) = -2e^{-2t}\)

Apply L’Hôpital:
[
\lim_{t \to 0} \frac{e^{-t} - 1}{e^{-2t} - 1} = \lim_{t \to 0} \frac{-e^{-t}}{-2e^{-2t}} = \lim_{t \to 0} \frac{e^{-t}}{2e^{-2t}} = \lim_{t \to 0} \frac{e^{t}}{2} = \frac{1}{2}.
]

So near time zero, the first object’s temperature difference is about half the second’s, in a relative sense.

If you’re curious about real cooling laws, Newton’s law of cooling is often discussed in university physics notes, such as those you might find through MIT OpenCourseWare.

Medication concentration change (a simple pharmacokinetics example)

A very simplified model for drug concentration in the bloodstream might use an exponential decay:
[
C(t) = C_0 e^{-kt}, ] where \(C_0\) is the initial concentration and \(k > 0\) is a rate constant.

Suppose you want to understand how the average concentration over a short time compares to the instantaneous concentration at time zero. One way to phrase that is:
[
\lim_{t \to 0} \frac{\frac{1}{t}\int_0^t C(s)\,ds}{C(0)}.
]

This is a bit advanced, but L’Hôpital’s Rule helps. The numerator is an integral, but by the Fundamental Theorem of Calculus, the derivative of
\(F(t) = \int_0^t C(s)\,ds\)
is just \(C(t)\).

Rewrite the limit as
[
\lim_{t \to 0} \frac{\int_0^t C(s)\,ds}{t C(0)}.
]

As \(t \to 0\), both numerator and denominator go to 0. Differentiate top and bottom with respect to \(t\):

• Top derivative: \(C(t)\)
• Bottom derivative: \(C(0)\)

Apply L’Hôpital:
[
\lim_{t \to 0} \frac{\int_0^t C(s)\,ds}{t C(0)} = \lim_{t \to 0} \frac{C(t)}{C(0)} = \frac{C(0)}{C(0)} = 1.
]

Interpretation: over a very short time interval, the average concentration is basically the initial concentration. This type of reasoning appears in pharmacology and medical research; for background on drug behavior in the body, you can explore resources from the U.S. National Library of Medicine.

Subtle forms: turning 1^∞, 0^0, and ∞^0 into quotients

Some of the best examples of L’Hôpital’s Rule don’t start as quotients. They look like weird power limits such as \(1^\infty\). To use L’Hôpital, we sneak in a logarithm.

Example: \(\lim_{x \to 0} (1 + x)^{1/x}\)

This is the famous limit that defines the number \(e\). Plug in \(x = 0\) informally and you get something like \(1^\infty\), which is indeterminate.

Set
[
y = (1 + x)^{1/x}.
]

Take natural logs:
[
\ln y = \frac{1}{x} \ln(1 + x).
]

Now consider the limit of \(\ln y\):
[
\lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{\ln(1 + x)}{x}.
]

As \(x \to 0\), numerator and denominator both go to 0, so we can use L’Hôpital:

• Top derivative: \(\frac{1}{1+x}\)
• Bottom derivative: 1

So
[
\lim_{x \to 0} \frac{\ln(1 + x)}{x} = \lim_{x \to 0} \frac{1}{1 + x} = 1.
]

Therefore \(\lim_{x \to 0} \ln y = 1\), which means [ \lim_{x \to 0} y = e^1 = e.
]

This is a textbook‑worthy example of L’Hôpital’s Rule where the original expression isn’t a fraction, but a clever logarithm turns it into one.

When not to use L’Hôpital’s Rule

Seeing all these examples of L’Hôpital’s Rule can make it tempting to use it everywhere. A few quick guardrails:

• Only use it when you truly have 0/0 or ∞/∞ after checking the limit.
• If simple algebra (factoring, canceling, rationalizing) works, that’s often faster and cleaner.
• Some limits (like oscillatory ones) can’t be fixed with derivatives alone.

Many instructors and university notes (for instance, those you might find via Harvard’s math department) emphasize: L’Hôpital’s Rule is powerful, but it’s not your only tool.

FAQ: Common questions about examples of L’Hôpital’s Rule

Q1: Can you give another quick example of L’Hôpital’s Rule with ∞/∞?
Yes. Consider
[
\lim_{x \to \infty} \frac{2x^2 + 3x}{5x^2 - 7}. ] Both numerator and denominator go to infinity, so it’s ∞/∞. Differentiate top and bottom: [ \lim_{x \to \infty} \frac{4x + 3}{10x}.
]
Still ∞/∞. Differentiate again:
[
\lim_{x \to \infty} \frac{4}{10} = \frac{2}{5}.
]
You could also do this by dividing top and bottom by \(x^2\), but this is a clean example of L’Hôpital’s Rule at work.

Q2: Are there examples of L’Hôpital’s Rule that produce infinity as an answer?
Yes. For instance,
[
\lim_{x \to 0^+} \frac{1}{x^2} ] blows up to \(\infty\), but that one doesn’t need L’Hôpital. A better example is [ \lim_{x \to 0^+} \frac{\ln x}{x - 1}.
]
As \(x \to 0^+\), \(\ln x \to -\infty\) and \(x - 1 \to -1\), so the whole fraction goes to \(+\infty\). L’Hôpital isn’t required, but you can confirm behavior by differentiating if you first rewrite the expression near \(x = 1\). The key point: L’Hôpital helps you clarify infinite behavior; it doesn’t force everything to be finite.

Q3: What’s a common mistake when using examples of L’Hôpital’s Rule on exams?
Two big ones: using it when the form isn’t 0/0 or ∞/∞, and forgetting to re‑evaluate the limit after differentiating. Always check the form first, then apply the rule, then check the new limit.

Q4: Do I always have to use L’Hôpital if I see 0/0?
No. Many 0/0 limits are easier with algebra or trigonometric identities. For example,
\(\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\) is faster by factoring. But having several examples of L’Hôpital’s Rule in your head helps when algebra gets messy.

Q5: Where can I study more real examples of limits and L’Hôpital’s Rule?
You can explore open course notes from universities such as MIT OpenCourseWare, or search for calculus resources through Harvard University and the National Library of Medicine. Many of these sources include worked examples and problem sets you can practice with.

If you keep these examples of L’Hôpital’s Rule—3 practical examples at the start plus the extra ones we explored—you’ll have a solid pattern library in your head. The more you practice recognizing 0/0 and ∞/∞ forms, the more automatic L’Hôpital’s Rule will feel, and the less intimidating those strange‑looking limits will become.