Limits and Rational Functions - Quick Reference

If a rational function has a common factor that vanishes at a point, you may have a removable discontinuity (a hole) at that point.
Example structure:
- Numerator: \((s+6)(s-2)\)
- Denominator: \((s+6)(s-6)\)
- For s ≠ -6, f(s) = \frac{(s+6)(s-2)}{(s+6)(s-6)} = \frac{s-2}{s-6}.
- At s = -6, the original form is 0/0, but the limit exists and equals \(\frac{-6-2}{-6-6} = \frac{-8}{-12} = \frac{2}{3}.\)
takeaway: compute limits by canceling common factors before substituting; keep limit notation to reflect approaching the point, not necessarily the function value at that point.

Steps:
1) Check if the numerator and denominator both vanish at the target value.
2) Factor to reveal common factors.
3) Cancel and evaluate the limit of the simplified expression.
Important: after cancellation, you evaluate the limit of the simplified expression, which gives the limit of the original function as x approaches the point (even if the function is undefined there).

When a zero/zero situation involves square roots in the numerator, multiply by a conjugate to create a difference of squares and simplify.
General idea:
- If numerator is \sqrt{A} - \sqrt{B}, multiply by \frac{\sqrt{A} + \sqrt{B}}{\sqrt{A} + \sqrt{B}} to obtain \frac{A - B}{(A - B) + other terms} form.
- Keep the overall value the same by multiplying by a form of one.
Note: do not multiply only the numerator; multiply by a form of one to keep the expression's value unchanged.

Infinite limit: occurs when the function grows without bound as x approaches c. The two-sided limit does not exist.
Oscillating limit: example \(\lim_{x\to 0} \sin(1/x)\) does not exist because the values oscillate between -1 and 1 without settling.
Squeeze (sandwich) theorem: if g(x) ≤ f(x) ≤ h(x) near c and \lim{x\to c} g(x) = \lim{x\to c} h(x) = L, then \lim_{x\to c} f(x) = L.
Application: for x near 0, |x| bounds x \sin(1/x); thus \lim_{x\to 0} x \sin(1/x) = 0 by squeezing.

Definition: the limit exists at c only if the left-hand limit (x → c^-) and the right-hand limit (x → c^+) agree.
Example: if \(\lim{x\to 1^-} f(x) = -2\) and \(\lim{x\to 1^+} f(x) = 1\), then \(\lim_{x\to 1} f(x)\) does not exist.
Important: a value at x = c may be defined differently from the limit; the two are not the same unless the function is continuous at c.

Keep limit notation when discussing approaching a point; do not replace with equality too early.
Graphing calculators may show holes (removable discontinuities) as undefined points due to pixel resolution; holes correspond to potential removable discontinuities.
L'Hôpital's Rule: mentioned as a check (not required for credit on quizzes/exams). It can verify answers but does not replace understanding of limits.
Last-minute exam strategy: understand the structure of the limit problem (factoring, conjugates, one-sided limits, and the squeeze theorem) rather than relying solely on memorized results.

Zero/zero can often be resolved by factoring and canceling common factors to reveal a finite limit.
If radicals are involved, consider multiplying by a conjugate to create a difference of squares.
If the limit is determined by left/right behavior, compute one-sided limits first.
Infinite and oscillating limits do not exist as two-sided limits; use squeeze theorem where helpful.
Always distinguish between the limit value and the function's value at the point of interest.