Selecting Procedures for Determining Limits
Introduced| English |
|---|
| flowchart/ˈfləʊtʃɑːt/ |
| procedure/prəˈsiːdʒə/ |
A limit is a "which tool?" puzzle
- You now own several ways to find a limit — the skill is picking the right one fast.
- Rushing into heavy algebra when substitution would work wastes time; guessing when it gives $\tfrac00$ wastes marks.
- Think of every limit as a short decision: look, try, then choose.
- This lesson is the flowchart 流程图 that ties the earlier tools together.
The decision flow
- Step 1 — try direct substitution. Plug $c$ in.
- If you get a real number, that is the limit. Done.
- If you get $\dfrac{0}{0}$ (indeterminate), go to Step 2.
- If you get $\dfrac{\text{nonzero}}{0}$, the size blows up — check for a vertical asymptote (an infinite limit or DNE), not algebra.
Probe a rational function
y = ax² + bx + c
Change the shape and ask whether plugging in gives a number, a hole ($\tfrac00$), or a blow-up ($\tfrac{n}{0}$) — that triage picks your tool.
Direct substitution into $\lim_{x\to 4}\dfrac{x-4}{x^2-16}$ gives $\tfrac00$. What next?
Polynomial over polynomial, both zero at $4$ → factor: $\frac{x-4}{(x-4)(x+4)}=\frac{1}{x+4}\to\frac18$.
The first procedure to attempt on almost any limit is direct ____.
It is fastest when it works and tells you which branch to take when it does not.
Step 2 — match the form to a technique
- Polynomial over polynomial, both zero at $c$ → factor and cancel.
- A square root causing the zero → rationalize with the conjugate.
- A difference of fractions → combine over a common denominator, then cancel.
- Special famous forms (like $\frac{\sin x}{x}$) → recognize the known limit or the Squeeze Theorem.
Substitution into $\lim_{x\to 3}\dfrac{x+1}{x-3}$ gives $\dfrac{4}{0}$. This signals...
Nonzero over zero blows up; factoring cannot help. The two-sided limit is DNE (one side $+\infty$, other $-\infty$).
Select all correct method-to-form matches.
The first three match. Nonzero over zero is an asymptote, not something factoring resolves.
Evaluate $\displaystyle\lim_{x\to 5}\dfrac{x^2-25}{x-5}$.
$\tfrac00$ → factor: $x+5 \to 10$.
Step 3 — justify and sanity-check
- State why your method is valid: "substitution works because the function is continuous here," or "cancelling is valid for $x\neq c$."
- Check the answer is consistent — does a quick table or the graph's shape agree?
- A good solution names its procedure 步骤 and confirms the result, not just a bare number.
- Efficiency comes from choosing well the first time, not from trying everything.
$\tfrac00$ and $\tfrac{5}{0}$ should be handled the same way.
$\tfrac00$ is indeterminate (simplify); $\tfrac{5}{0}$ blows up (asymptote). Different situations, different responses.
Do not confuse the two "over zero" outcomes. $\dfrac00$ is indeterminate → simplify with algebra. But $\dfrac{5}{0}$ (nonzero over zero) is not indeterminate → the magnitude grows without bound, so you are looking at an infinite limit / vertical asymptote, and factoring will not help.
Choose a method for each, then evaluate:
- $\displaystyle\lim_{x\to 2}(x^2+1)$: substitution → $5$. (Continuous — no tricks.)
- $\displaystyle\lim_{x\to 5}\dfrac{x^2-25}{x-5}$: substitution gives $\tfrac00$ → factor → $x+5 \to 10$.
- $\displaystyle\lim_{x\to 3}\dfrac{1}{x-3}$: substitution gives $\tfrac{1}{0}$ → not algebra; the one-sided limits are $\pm\infty$, so the two-sided limit is DNE.
Selecting a procedure is a quick flow: substitute first. A real number is the answer; $\tfrac00$ means simplify (factor / rationalize / combine, or a known limit); $\tfrac{\text{nonzero}}{0}$ means an infinite limit or asymptote, not algebra. Then justify the choice and sanity-check the result.