Determining Limits Using Algebraic Properties of Limits
Introduced| English | Meaning |
|---|---|
| limit laws/ˈlɪmɪt lɔːz/ | Rules for combining limits of sums, products, quotients and other expressions when their required conditions hold. |
| direct substitution/daɪˈrekt ˌsʌbstɪˈtjuːʃn/ | Evaluating an expression by replacing its variables with the given values. |
| polynomial/ˌpɒlɪˈnəʊmɪəl/ | An expression formed by adding terms with non-negative integer powers of its variables. |
| rational/ˈræʃənl/ | Describing a number expressible as a ratio of two integers with a nonzero denominator, including a fractional exponent of this form. |
| indeterminate form/ˌɪndɪˈtɜːmɪnət fɔːm/ | A limiting expression whose apparent form alone does not determine its limit, such as zero over zero. |
Limits obey friendly algebra
- Once you know the limits of two functions, you can combine them like ordinary numbers.
- Whatever $f$ and $g$ head toward, their sum heads toward the sum of those targets.
- The same works for differences, products, and (carefully) quotients.
- These are the limit laws 极限法则 — they let you break a messy limit into simple pieces.
The limit laws, in one place
- Suppose $\displaystyle\lim_{x\to c} f(x)=L$ and $\displaystyle\lim_{x\to c} g(x)=M$. Then:
- Sum / difference: $\lim (f\pm g) = L\pm M$.
- Product: $\lim (f\cdot g) = L\cdot M$; constant multiple: $\lim (k f)=kL$.
- Quotient: $\displaystyle\lim \frac{f}{g}=\frac{L}{M}$, provided $M\neq 0$.
Given $\lim f = 6$ and $\lim g = 3$, find $\lim \dfrac{f}{g}$.
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Quotient law (denominator limit $3\neq0$): $\frac{6}{3}=2$.
The quotient law $\lim\frac fg=\frac LM$ requires that $M$, the denominator's limit, is not ____.
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If $M=0$ the law does not apply directly — you may have an indeterminate form.
Direct substitution: the fast lane
- For a polynomial 多项式 or a rational 有理 function, the laws add up to one shortcut: just plug $c$ in.
- $\displaystyle\lim_{x\to 3}(x^2-2x+1) = 3^2-2(3)+1 = 4$ — no table, no graph.
- Direct substitution 直接代入 works whenever the function is defined and continuous at $c$.
- This is always the first thing to try — it is fastest when it works.
A polynomial has no surprises
y = ax² + bx + c
Polynomials are smooth everywhere — drag the coefficients and see why plugging in $c$ (direct substitution) simply reads off the height.
Evaluate $\displaystyle\lim_{x\to 2}(3x^2 - x + 4)$.
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Direct substitution: $3(4)-2+4 = 14$.
For which limits does direct substitution give the answer immediately?
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The third gives $\tfrac00$ (denominator limit $0$) so substitution stalls; the others give a finite value straight away.
When substitution stalls
- Plug in and you might hit $\dfrac{0}{0}$: an indeterminate form 不定型.
- Indeterminate does not mean the limit fails — it means substitution alone can't decide, and more work is needed (next lesson).
- Beware the quotient law's fine print: it needs $M\neq0$. If the denominator's limit is $0$, the law does not apply directly.
- So: try substitution first; if it gives a real number, you are done; if it gives $\tfrac00$, switch strategies.
Direct substitution gives $\tfrac00$. This means...
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$\tfrac00$ is indeterminate: a cue to do algebra, not a final answer.
When you meet a new limit, direct substitution is the sensible first thing to try.
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It is fastest when it works; only switch strategies if it yields an indeterminate form.
$\dfrac{0}{0}$ is indeterminate, not "zero" and not "undefined-so-DNE." It is a signal that the limit is hiding and you must simplify first. Do not report $\tfrac00$ as an answer, and do not conclude the limit does not exist — $\lim_{x\to2}\frac{x^2-4}{x-2}$ is $\tfrac00$ on substitution yet equals $4$.
Find $\displaystyle\lim_{x\to 1}\dfrac{2x^2+3}{x+4}$.
- Numerator limit: $2(1)^2+3 = 5$. Denominator limit: $1+4 = 5$.
- The denominator's limit is $5\neq 0$, so the quotient law applies.
- $\displaystyle\lim_{x\to 1}\dfrac{2x^2+3}{x+4} = \dfrac{5}{5} = 1$ by direct substitution.
The limit laws let you split a limit across sums, differences, products, and quotients (quotient needs a non-zero denominator limit). For polynomials and rationals this collapses to direct substitution — always try it first. A result of $\tfrac00$ is an indeterminate form: not an answer, but a cue to simplify.