Unit 4 Differentiation Applications: Approximations and Indeterminate Limits

Local Linearity and Approximation

What “local linearity” means

A major idea behind derivatives is that even when a function is curved overall, it looks almost like a straight line if you zoom in close enough to a point. This is called local linearity. The derivative at a point measures the slope of the line that best matches the function right near that point.

That “best matching” line is the tangent line. If you know the tangent line at a nearby, convenient point, you can use that line to estimate function values that might otherwise be hard to compute exactly.

This matters because many real problems involve small changes—slight increases in temperature, tiny changes in time, small manufacturing tolerances, small measurement errors. Linear approximations turn complicated functions into simple arithmetic when the change is small.

From tangent line to linearization

Suppose you have a differentiable function f and you pick an input value a where you know (or can compute) f(a) and f'(a). The tangent line to y=f(x) at x=a is

y = f(a) + f'(a)(x-a)

The expression on the right is so useful that we give it a name. The linearization of f at x=a is the function L defined by

L(x) = f(a) + f'(a)(x-a)

Conceptually:

f(a) sets the “anchor point” (the line passes through \big(a,f(a)\big)).
f'(a) sets the slope.
x-a is how far you moved from the anchor input.

When x is close to a, you typically have

f(x) \approx L(x)

This is the mathematical version of “zoom in and it’s basically straight.”

Why linearization is an approximation (and when it’s good)

Linearization is not magic—it’s an approximation. It works well when x is close to a because the function’s curvature doesn’t have much time to pull away from the tangent line.

A useful mental picture: the tangent line “kisses” the curve at x=a and then starts to drift away. The farther you move, the more the curve’s concavity matters.

Two practical guidelines:

Choose a close to the x you care about.
Choose a where f(a) and f'(a) are easy to compute.

Notation you’ll see (and what it means)

Linear approximation problems sometimes use slightly different notations for the same idea.

Idea	Common notation	Meaning
Linearization	L(x)	Tangent-line function used to approximate f(x) near a
Tangent line at a	y = f(a)+f'(a)(x-a)	Same as y=L(x)
Small change in input	\Delta x	New input is a+\Delta x
Small change in output	\Delta y = f(a+\Delta x)-f(a)	Actual change in function value

You’ll also see a “differentials” version (next section), which is another way to track small changes.

Using linearization step by step

A typical approximation task is: estimate f(b) for some b close to a.

Process:

Pick a nearby point a where the function is easy.
Compute f(a).
Compute f'(a).
Write L(x)=f(a)+f'(a)(x-a).
Approximate f(b) by evaluating L(b).

Example 1: Approximating a square root

Estimate \sqrt{50} using a linearization.

Why linearization helps: Square roots are harder to do by hand, but they come from the function f(x)=\sqrt{x}, whose derivative is manageable.

f(x)=\sqrt{x}

Choose a=49 because \sqrt{49}=7 and 50 is close to 49.
Compute f(a):

f(49)=7

Compute f'(x):

f'(x)=\frac{1}{2\sqrt{x}}

f'(49)=\frac{1}{2\cdot 7}=\frac{1}{14}

Build the linearization:

L(x)=7+\frac{1}{14}(x-49)

Evaluate at x=50:

L(50)=7+\frac{1}{14}(1)=7+\frac{1}{14}=7.071428\ldots

\sqrt{50} \approx 7.0714

Common sense check: since 50 is slightly bigger than 49, the square root should be slightly bigger than 7. That matches.

Example 2: Estimating a trig value near a familiar angle

Estimate \sin(0.1) (radians).

f(x)=\sin x

Choose a=0 because \sin 0 = 0 and 0.1 is close to 0.
Compute f(a):

f(0)=0

Compute f'(x):

f'(x)=\cos x

f'(0)=1

Linearization:

L(x)=0+1(x-0)=x

Approximate:

\sin(0.1) \approx L(0.1)=0.1

This is a famous local linearity result: near 0, \sin x behaves like x.

Differentials: a “small change” language for approximation

Linearization is often presented alongside differentials, which package the same tangent-line idea into a compact way to estimate changes.

If y=f(x), the differential relationship is

dy = f'(x)\,dx

Here’s how to interpret it in AP Calculus AB context:

dx is a small change in the input (often chosen as dx=\Delta x when the change is small).
dy is the estimated change in the output produced by the tangent line.

So if you start at x=a and change by \Delta x, you can estimate

\Delta y \approx dy = f'(a)\Delta x

This is the same idea as linearization, because

L(a+\Delta x) = f(a) + f'(a)\Delta x

and therefore

L(a+\Delta x)-f(a)=f'(a)\Delta x

Example 3: Estimating the effect of measurement error

Suppose the radius of a sphere is measured as r=10 cm, but the measurement could be off by 0.1 cm. Estimate the resulting error in computed volume.

The volume is

V=\frac{4}{3}\pi r^3

Differentiate with respect to r:

\frac{dV}{dr}=4\pi r^2

A small radius error dr produces an estimated volume error

dV \approx 4\pi r^2\,dr

At r=10 and dr=0.1:

dV \approx 4\pi (10)^2(0.1)=4\pi\cdot 100\cdot 0.1=40\pi

So the computed volume could be off by about

40\pi \text{ cm}^3

The key takeaway is not the exact number—it’s the method: derivatives translate small input uncertainty into estimated output uncertainty.

What can go wrong with linear approximations

Local linearity is powerful, but several predictable mistakes show up:

Using a point a that is not close. If b is far from a, the curve’s concavity can make the tangent line a poor model.
Forgetting the “anchor shift” x-a. Students sometimes write L(x)=f(a)+f'(a)x, which only works when a=0.
Mixing up actual change and estimated change. \Delta y is the actual change; dy is the linear (tangent-line) estimate.

Exam Focus

Typical question patterns:
- “Use the tangent line at x=a to approximate f(b),” sometimes with a calculator-prohibited value like a square root or trig value.
- “Find the linearization L(x) of f at x=a and use it to estimate …”
- “Use differentials to estimate the error/percent error in a measurement-based computation.”
Common mistakes:
- Computing f'(a) correctly but plugging into f'(x) instead of evaluating at a.
- Choosing a convenient a but not checking that b is sufficiently close.
- Treating dy as exact (or reporting too many digits) instead of as an estimate.

L'Hôpital's Rule

Why indeterminate forms appear in limits

In calculus, many important limits involve expressions that look like they should be straightforward to evaluate by substitution, but substitution produces an ambiguous signal.

For example, if you try to compute

\lim_{x\to 0}\frac{\sin x}{x}

direct substitution gives

\frac{\sin 0}{0} = \frac{0}{0}

The expression \frac{0}{0} is not a number and it doesn’t tell you what the limit is. It’s called an **indeterminate form** because different functions can produce the same “surface appearance” \frac{0}{0} while having different limit values.

Another common indeterminate form is

\frac{\infty}{\infty}

which happens when both numerator and denominator grow without bound but at potentially different rates.

L'Hôpital's Rule is a technique that often resolves these limits by replacing a difficult ratio with a new ratio involving derivatives.

What L'Hôpital's Rule says (core AB version)

L'Hôpital's Rule applies to limits of a quotient when you have one of these indeterminate forms:

\frac{0}{0}
\frac{\infty}{\infty}

Informally: if you have a quotient \frac{f(x)}{g(x)} and as x\to a both f(x) and g(x) go to 0 (or both go to \pm\infty), then under appropriate differentiability conditions you can differentiate top and bottom and try again.

In practice, the version you use is:

\lim_{x\to a}f(x)=0 \text{ and } \lim_{x\to a}g(x)=0

\lim_{x\to a}f(x)=\pm\infty \text{ and } \lim_{x\to a}g(x)=\pm\infty

then you can try

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

provided the right-hand limit exists (or is infinite) and the functions are differentiable near a (except possibly at a itself).

How it works (intuition)

L'Hôpital's Rule is deeply connected to the idea of local linearity:

Near x=a, if f(a)=0 and g(a)=0, the first-order behavior of f and g is controlled by their derivatives.
Roughly, near a you can think of

f(x) \approx f'(a)(x-a)

and

g(x) \approx g'(a)(x-a)

So the ratio behaves like

\frac{f(x)}{g(x)} \approx \frac{f'(a)(x-a)}{g'(a)(x-a)} = \frac{f'(a)}{g'(a)}

That’s not a proof, but it explains why “differentiate numerator and denominator” can reveal the hidden value.

The correct procedure (and the most common pitfall)

Step 1: Try substitution first.
If substitution gives a real number, you’re done.

Step 2: Verify an indeterminate form.
You must explicitly get \frac{0}{0} or \frac{\infty}{\infty}. If you get something like \frac{0}{5} or \frac{\infty}{7}, that is not indeterminate.

Step 3: Apply L'Hôpital once (differentiate top and bottom).
Differentiate f and g separately:

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

Step 4: Re-check the new limit.

If substitution now works, finish.
If you still get \frac{0}{0} or \frac{\infty}{\infty}, you may apply L'Hôpital again.
If you get a different indeterminate form, you may need algebra first.

Most common pitfall: Applying L'Hôpital without confirming \frac{0}{0} or \frac{\infty}{\infty}. On AP-style solutions, that missing check is often treated as a reasoning error.

Examples of L'Hôpital's Rule

Example 1: Classic \frac{0}{0} trig limit

Evaluate

\lim_{x\to 0}\frac{\sin x}{x}

Substitution:

\frac{\sin 0}{0} = \frac{0}{0}

Indeterminate, so L'Hôpital applies.

Differentiate numerator and denominator:

\lim_{x\to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{\cos x}{1}

Substitute:

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

So the limit is 1.

Example 2: Exponential minus 1 (common AB pattern)

Evaluate

\lim_{x\to 0}\frac{e^x-1}{x}

Substitution:

\frac{e^0-1}{0} = \frac{0}{0}

Apply L'Hôpital:

\lim_{x\to 0}\frac{e^x-1}{x} = \lim_{x\to 0}\frac{e^x}{1}

Substitute:

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

So the limit is 1.

Example 3: A polynomial “infinity over infinity” growth comparison

Evaluate

\lim_{x\to \infty}\frac{3x^2+1}{5x^2-7x}

As x\to\infty, numerator \to\infty and denominator \to\infty, so you have \frac{\infty}{\infty}.
Apply L'Hôpital once:

\lim_{x\to \infty}\frac{3x^2+1}{5x^2-7x} = \lim_{x\to \infty}\frac{6x}{10x-7}

Still \frac{\infty}{\infty}, apply again:

\lim_{x\to \infty}\frac{6x}{10x-7} = \lim_{x\to \infty}\frac{6}{10}

So the limit is

\frac{3}{5}

(You could also do this limit by dividing by x^2, but L'Hôpital is a valid method when the indeterminate form is verified.)

Other indeterminate forms and how to handle them

L'Hôpital's Rule directly addresses only \frac{0}{0} and \frac{\infty}{\infty}. But AP problems may present other indeterminate forms that you can convert into one of those two.

Common ones:

0\cdot\infty
\infty-\infty
0^0, 1^\infty, \infty^0 (these are typically handled by rewriting using logarithms; whether this appears depends on course scope and problem design, but the conversion idea is good to recognize)

Converting 0\cdot\infty

Example structure:

\lim_{x\to 0^+} x\ln x

As x\to 0^+, \ln x\to -\infty and x\to 0, so this looks like 0\cdot (-\infty), an indeterminate form.

To use L'Hôpital, rewrite as a fraction. A common move is to put one factor in the denominator:

x\ln x = \frac{\ln x}{1/x}

Now as x\to 0^+, \ln x\to -\infty and 1/x\to \infty, giving \frac{-\infty}{\infty}, which is an allowed L'Hôpital form.

Apply L'Hôpital:

\lim_{x\to 0^+}\frac{\ln x}{1/x} = \lim_{x\to 0^+}\frac{1/x}{-1/x^2}

Simplify the expression:

\frac{1/x}{-1/x^2} = -x

Now evaluate:

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

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

What goes wrong here: A frequent error is differentiating x\ln x directly with the product rule and trying to “plug in” x=0. But \ln x is not defined at 0, and you’re doing a limit, not evaluating a function at a point.

Converting \infty-\infty

Example structure:

\lim_{x\to \infty}(\sqrt{x^2+x}-x)

Both terms go to \infty, so this is \infty-\infty (indeterminate). The standard fix is algebra, often using a conjugate.

Multiply by the conjugate:

\sqrt{x^2+x}-x = \frac{(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)}{\sqrt{x^2+x}+x}

The numerator becomes a difference of squares:

(x^2+x)-x^2 = x

So the expression is

\frac{x}{\sqrt{x^2+x}+x}

Now as x\to\infty, you have \frac{\infty}{\infty}, so L'Hôpital can be used if desired. Often, it’s easier to factor x out of the square root:

\sqrt{x^2+x} = x\sqrt{1+1/x}

Then

\frac{x}{\sqrt{x^2+x}+x} = \frac{x}{x\sqrt{1+1/x}+x} = \frac{1}{\sqrt{1+1/x}+1}

Now substitute x\to\infty so 1/x\to 0:

\frac{1}{\sqrt{1+0}+1} = \frac{1}{2}

So the limit is \frac{1}{2}.

Why this matters: Many students try L'Hôpital immediately on \sqrt{x^2+x}-x, but L'Hôpital needs a quotient, not a difference. The algebra step is the key.

Connecting L'Hôpital to local linearity

It’s not an accident that linearization and L'Hôpital show up near each other: both rely on the idea that derivatives describe local behavior.

Linearization approximates a function near a point with a tangent line.
L'Hôpital approximates a ratio of functions near a point by comparing their derivatives.

In both cases, the derivative is acting like the “first-order” description of how the function behaves under tiny changes.

What can go wrong with L'Hôpital (beyond the basic pitfall)

Applying it to non-indeterminate forms. For instance, if substitution gives \frac{0}{2}, the limit is 0; L'Hôpital is unnecessary and can even lead you into undefined expressions.
Forgetting to re-check after differentiating. You must evaluate the new limit. Sometimes one application is enough; sometimes you need more; sometimes the limit diverges.
Differentiation errors under pressure. Because L'Hôpital replaces one limit with another, a small derivative mistake can change the entire outcome.
Ignoring domain issues. For one-sided limits like x\to 0^+ with \ln x, the direction matters.

Exam Focus

Typical question patterns:
- “Evaluate the limit” where substitution produces \frac{0}{0} or \frac{\infty}{\infty}, often involving trig, exponentials, logarithms, or rational functions.
- Limits that are not initially a quotient but can be rewritten into one (products like 0\cdot\infty or differences like \infty-\infty).
- Free-response solutions that expect you to show the indeterminate form before using L'Hôpital.
Common mistakes:
- Skipping the indeterminate-form verification step and losing justification points.
- Differentiating the denominator incorrectly (especially with chain rule expressions like 1/x or \sqrt{\cdot}).
- Using L'Hôpital when algebra or known limits would be simpler, then getting stuck after creating a more complicated expression.