The Limit Laws and Indeterminate Forms
Apply finite limit laws where their conditions hold. Investigate 0/0 forms with suitable algebra or bounds; the form alone guarantees neither a removable factor nor a finite limit.
What a learner can do afterwards
- Apply finite limit laws and investigate indeterminate quotients using valid algebra
- Evaluate root-difference limits by rationalising and checking the remaining expression
- Use bounding arguments and radian trigonometric limits, including rescaling
1 · Read
If f and g have finite limits L and M, their sum has limit L+M and their product has limit LM. Their quotient has limit L/M only when M is nonzero. Constants keep their values. For example, if the component limits are 2 and 4, adding one to their product gives limit 2(4)+1 = 9.
A 0/0 result from substitution is indeterminate. It does not decide whether the limit is finite, infinite or absent. A counterexample to automatic nonexistence is x/x: away from zero it equals 1. A common factor need not remove every singularity, as x/x² leaves 1/x.
For (x²-25)/(x-5) near 5, factor the numerator as (x-5)(x+5). Cancel only for x ≠ 5. The nearby expression becomes x+5 and has limit 10. Equal expressions on a punctured neighborhood have the same limit; the expression need not be defined at the approach point.
For cubes, use x³-a³ = (x-a)(x²+ax+a²) and x³+a³ = (x+a)(x²-ax+a²). If the denominator contains the matching linear factor, cancel it where nonzero. Evaluate the remaining polynomial at the approach point. Check what remains instead of assuming cancellation guarantees a finite limit.
For (sqrt(x+16)-4)/x near 0, multiply top and bottom by sqrt(x+16)+4. The numerator becomes x. For x ≠ 0, cancellation leaves 1/(sqrt(x+16)+4), whose limit is 1/8. The identity (a-b)(a+b) = a²-b² also works when a root difference is in the denominator.
The squeeze theorem requires lower and upper bounds near the point with the same limit. If A-(x-c)² ≤ g(x) ≤ A+(x-c)², then g tends to A at c. Bounds with different limits do not establish a unique limit. For instance, bounds -2 and 2 alone allow several constant functions.
A bounded factor can be controlled by a vanishing amplitude. If |b(x)| ≤ M near c for a fixed M ≥ 0 and a(x) tends to 0, then |a(x)b(x)| ≤ M|a(x)| tends to 0. The product therefore tends to 0. Sine is bounded by 1 in magnitude, even when its argument keeps changing. Adding a constant shifts the product's limit by that constant. The bounds establish the result without substituting into an oscillating factor.
Choose a method from the actual expression, not from 0/0 alone. Factoring or conjugates may give an equivalent expression; inspect its remaining limit. Bounds may establish a limit directly. Neither cancellation nor squeezing promises that final substitution into the original expression will work.
A radian angle is signed arc length divided by radius. On the unit circle, cos u and sin u are the horizontal and vertical coordinates; tan u = sin u/cos u where cos u ≠ 0. Cosine is continuous and cos 0 = 1. For 0 < |u| < pi/2, cos u ≤ sin u/u ≤ 1, so sin u/u tends to 1. For fixed k ≠ 0, sin(kx)/x = k sin(kx)/(kx) tends to k. Also tan(kx)/x = [sin(kx)/(kx)][k/cos(kx)] tends to k.
A 0/0 form needs investigation. Valid algebra can preserve nearby values; matching bounds can establish a limit directly.
Original limit-law lesson, questions and flashcards written for this learning dataset.
Practice questions written for this learning dataset. The rewritten readings were written for this dataset.
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