Practice: Limits done carefully

Self-check

Six short questions. Answer each one in your head (or on paper) before opening the collapsible. Trying to retrieve the answer is where the learning sticks; rereading feels productive but does much less.

1. For which limits can you just “plug in,” and which are the interesting ones?

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For continuous (well-behaved) functions, the limit equals the value: lim (x->2) x² = 4, just plug in. The interesting cases are indeterminate forms (0/0, ∞/∞), where plugging in fails and gives something meaningless. The derivative is the headline example, since (f(x+h)-f(x))/h becomes 0/0 at h = 0.

2. State the plain-English meaning of lim (x->a) f(x) = L.

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You can force f(x) as close to L as you want, just by making x close enough to a. Formally (epsilon-delta): for every output precision ε > 0 you demand, there is an input window δ > 0 around a such that any x within δ of a (but not equal to it) lands within ε of L. Challenge (any ε), response (a δ that delivers it).

3. When does L’Hôpital’s rule apply, and what does it say?

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It applies only when lim f(x)/g(x) has the indeterminate form 0/0 or ∞/∞. Then lim f(x)/g(x) = lim f'(x)/g'(x): differentiate the numerator and denominator separately (not the quotient rule) and try the limit again. Always check the form first; on a determinate value, L’Hôpital does not apply.

4. What do you do if one application of L’Hôpital still leaves 0/0?

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Apply the rule again, and again, as long as the indeterminate form persists. For example, (1 - cos x)/x² goes to sin x/(2x) (still 0/0), then to cos x/2 = 1/2. Stop only when plugging in gives a determinate value.

5. Why does L’Hôpital’s rule work?

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Near a, each function is approximately its value plus its slope times the displacement: f(x) ≈ f(a) + f'(a)·(x-a). In the 0/0 case f(a) = g(a) = 0, so the constant terms vanish and f/g ≈ f'(a)(x-a)/[g'(a)(x-a)] = f'(a)/g'(a): the shared (x-a) cancels, leaving the ratio of derivatives. It keeps only the leading behavior that determines where the ratio heads. (This is the first-order Taylor idea.)

6. Give an example of a limit that does not exist, and how epsilon-delta confirms it.

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sin(1/x) as x -> 0: as x shrinks, 1/x races to infinity, so sin(1/x) swings between -1 and 1 infinitely fast, never settling. Epsilon-delta confirms it: pick ε = 0.5; no matter how small a window δ you choose around 0, the function still takes values near +1 and near -1 inside it, so it cannot stay within 0.5 of any candidate L. No limit exists.

Try it yourself, part 1: apply L’Hôpital

Pen and paper, about 7 minutes. Each limit is indeterminate at first; apply L’Hôpital (differentiating top and bottom separately), repeating if the form persists.

(a) lim (x->0) (e^(2x) - 1)/x

(b) lim (x->0) sin(3x)/x

(c) lim (x->0) (x - sin x)/x³

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(a) e^0 - 1 = 0, so it is 0/0. Differentiate top (2e^(2x), chain rule) and bottom (1):

lim (x->0) (e^(2x)-1)/x = lim (x->0) 2e^(2x)/1 = 2·e^0 = 2

(b) sin 0 = 0, so 0/0. Top derivative 3cos(3x) (chain rule), bottom 1:

lim (x->0) sin(3x)/x = lim (x->0) 3cos(3x)/1 = 3·cos 0 = 3

(c) 0/0. This one needs three passes:

(x - sin x)/x³   ->  (1 - cos x)/(3x²)   (still 0/0)
                 ->  sin x/(6x)           (still 0/0)
                 ->  cos x/6   ->  cos 0/6 = 1/6

The answer is 1/6. Keep applying the rule while the form stays indeterminate; stop when a plug-in finally works.

Try it yourself, part 2: an epsilon-delta challenge-response

About 4 minutes. Show that lim (x->3) 2x = 6 by finding, for a demanded precision ε, a window δ that delivers it.

Steps. (1) Write the output gap |2x - 6| in terms of |x - 3|. (2) Find the δ (in terms of ε) that forces the gap below ε. (3) Give the δ for ε = 0.1 and for ε = 0.001.

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|2x - 6| = 2·|x - 3|

To force 2·|x - 3| < ε, keep |x - 3| < ε/2. So choosing δ = ε/2 works: any x within ε/2 of 3 lands within ε of 6.

For ε = 0.1: δ = 0.05.
For ε = 0.001: δ = 0.0005.

Tighten the demanded precision and the window shrinks to match, exactly. That challenge-and-response, hand me any ε, I hand back a δ that delivers it, is what it means for the limit to be 6. (Here the function is a clean line, so δ = ε/2 is exact; for a curve like x² you also have to bound the wandering factor, as the lesson did.)

Flashcards

Nine cards. Click any card to reveal the answer. Use the Print flashcards button to lay out the full set as one card per page, ready to print or save as a PDF for offline review.

Q. When can you find a limit by just plugging in?

For continuous (well-behaved) functions, where the limit equals the value: lim (x->2) x² = 4. The interesting cases are indeterminate forms (0/0, ∞/∞), where plug-in fails.

Q. What does lim (x->a) f(x) = L mean, in plain English?

You can force f(x) as close to L as you want by making x close enough to a. Formally (epsilon-delta): for any precision ε, there is a window δ around a such that x within δ keeps f(x) within ε of L.

Q. What is an indeterminate form?

An expression like 0/0 or ∞/∞ that could approach any value (or none) depending on how the top and bottom shrink or grow. Plugging in gives no answer; you need a tool like L’Hôpital’s rule.

Q. State L'Hôpital's rule and its conditions.

If lim f(x)/g(x) is 0/0 or ∞/∞, then it equals lim f'(x)/g'(x): differentiate numerator and denominator separately (not the quotient rule), then retry. Only for those indeterminate forms; check the form first.

Q. Compute lim (x->0) sin(x)/x with L'Hôpital.

It is 0/0. Differentiate top (cos x) and bottom (1): lim cos(x)/1 = cos 0 = 1. This proves the small-angle approximation sin(x) ≈ x from the trig lesson.

Q. What if one application of L'Hôpital still gives 0/0?

Apply it again, repeating while the form stays indeterminate. Example: (1 - cos x)/x² -> sin x/(2x) -> cos x/2 = 1/2. Stop when a plug-in gives a determinate value.

Q. Why does L'Hôpital's rule work?

Near a, f(x) ≈ f(a) + f'(a)(x-a). In the 0/0 case the constant terms vanish, so f/g ≈ f'(a)(x-a)/[g'(a)(x-a)] = f'(a)/g'(a): the shared (x-a) cancels, leaving the ratio of derivatives (the leading behavior).

Q. How does epsilon-delta show sin(1/x) has no limit as x->0?

sin(1/x) swings between -1 and 1 infinitely fast near 0. Pick ε = 0.5: any window δ around 0 still contains values near +1 and near -1, so the function cannot stay within 0.5 of any L. No limit exists.

Q. Why do limits matter in machine learning?

They underwrite the field’s guarantees: convergence analysis (does gradient descent settle?), continuous-time models (the step-size -> 0 limit of an ODE solver), and universal approximation theorems (limit-based existence). Rarely computed by hand, always underneath.