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Linear Approximation & L'Hopital's Rule

Two everyday payoffs of the [[derivative-calculus|derivative]]. First, [[linear-approximation|linear approximation]]: zoom in on any smooth curve and it looks like a straight line, which lets you estimate messy values like sqrt(4.1) in your head. Second, [[lhopitals-rule|L'Hopital's rule]]: when a [[limit|limit]] collapses to a meaningless 0/0, derivatives can rescue it.

Zoom in and the curve goes straight

Pick any point on a smooth curve and keep zooming in on it. The bend gets gentler and gentler until, close enough, the curve is indistinguishable from a straight line — and that line is exactly the tangent line you met when you first learned the derivative. This is the whole idea behind linear approximation: near a point a, the complicated function f behaves almost like its tangent. The slope of that tangent is f'(a), so the line passing through the point (a, f(a)) is L(x) = f(a) + f'(a)(x - a). For x close to a, f(x) is approximately L(x).

Estimating sqrt(4.1) by hand

Say you need sqrt(4.1) and have no calculator. The exact answer is awkward, but sqrt(4) = 2 is easy, and 4.1 sits right next to 4. So let f(x) = sqrt(x) and anchor at a = 4. The derivative is f'(x) = 1/(2 sqrt(x)), so f'(4) = 1/(2 times 2) = 1/4. Now just walk the tangent line a tiny step of x - a = 0.1.

f(x)  = sqrt(x),   a = 4,   x = 4.1
f(a)  = sqrt(4) = 2
f'(a) = 1/(2 sqrt(4)) = 1/4

sqrt(4.1) ~ f(a) + f'(a)(x - a)
          = 2 + (1/4)(0.1)
          = 2.025

(true value 2.0248...  -> off by about 0.0002)
The estimate 2.025 is correct to three decimal places — from arithmetic you can do in your head.

There is a tidy notation for the rise alone. The little change in x is called dx and the resulting change in the tangent's height is the differential dy = f'(a) dx. Here dy = (1/4)(0.1) = 0.025, exactly the bump we added to 2. Differentials are just a compact way of saying 'slope times step', which is why dy/dx and f'(x) mean the same thing.

L'Hopital's rule rescues 0/0

Now the second payoff. Some limits hand you a ratio where both top and bottom rush to 0 at once, like lim x->0 sin(x)/x. Plugging in gives 0/0, which is not 0, not 1, not infinity — it is an indeterminate form, a question the symbols alone cannot answer. The same trouble appears as infinity/infinity. L'Hopital's rule says: when a limit of f(x)/g(x) is genuinely 0/0 or infinity/infinity, you may differentiate top and bottom separately and take the limit of f'(x)/g'(x) instead.

  1. Check the form first. For lim x->0 sin(x)/x, substitute x = 0: sin(0)/0 = 0/0. Yes, it is indeterminate, so the rule is allowed.
  2. Differentiate top and bottom on their own (NOT the quotient rule): d/dx sin(x) = cos(x), d/dx x = 1.
  3. Take the new limit: lim x->0 cos(x)/1 = cos(0)/1 = 1. So lim x->0 sin(x)/x = 1.

Use it only when the form earns it

L'Hopital is powerful, which makes it easy to misuse. The single most common mistake is applying it to a limit that is NOT indeterminate. Take lim x->0 cos(x)/(x + 1): substituting gives 1/1 = 1, a perfectly good answer — but differentiating top and bottom would give -sin(x)/1, whose limit is 0, the wrong answer. Always confirm you really have 0/0 or infinity/infinity before reaching for the rule.

Both ideas in this rung lean on the same insight: near a point, a smooth function is well described by its slope. Linear approximation uses that slope to predict a nearby value; L'Hopital uses the slopes of top and bottom to settle a ratio that the raw values left undecided. Same derivative, two very different rescues.