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← Everything for Calculus I
Calculus ILimits and continuity

Limit laws, and what to do when they fail.

Learn when you can just plug in, how to rescue a 0/0, and how to trap a wild function. Then switch to Solve and work through every class question.

The limit laws · Step 1 of 5

Break a limit into pieces you already know

The limit laws (when both limits exist)
  • Sum
    lim⁡ [f+g]=lim⁡f+lim⁡g{\displaystyle \lim\,[f + g] } = {\displaystyle \lim f } + {\displaystyle \lim g}
  • Difference · used here
    lim⁡ [f−g]=lim⁡f−lim⁡g{\displaystyle \lim\,[f - g] } = {\displaystyle \lim f } - {\displaystyle \lim g}
  • Constant multiple · used here
    lim⁡ [c f]=clim⁡f{\displaystyle \lim\,[c\,f] } = {\displaystyle c \lim f}
  • Product
    lim⁡ [f g]=lim⁡f⋅lim⁡g{\displaystyle \lim\,[f\,g] } = {\displaystyle \lim f \cdot \lim g}
  • Quotient
    lim⁡fg=lim⁡flim⁡gif lim⁡g≠0{\displaystyle \lim \frac{f}{g} } = {\displaystyle \frac{\lim f}{\lim g} \quad \text{if } \lim g } \ne {\displaystyle 0}
  • Power
    lim⁡ [f]n=[lim⁡f]n{\displaystyle \lim\,[f]^n } = {\displaystyle [\lim f]^n}
If lim⁡f{\lim f} and lim⁡g{\lim g} both exist, a limit passes straight through +{+}, −{-}, ×{\times}, powers and constants. Division works too, as long as the bottom’s limit is not 0. Each of these rules is a limit law, and on a test you may be asked to name the ones you used.
Try it: This is the class example: lim⁡x→5f(x)=4{\lim\limits_{x \to 5} f(x) } = { 4} and lim⁡x→5g(x)=−3{\lim\limits_{x \to 5} g(x) } = { -3}. Pick an expression and work out the answer before you open the working. Then slide lim⁡g{\lim g} to 0 and try the quotient.
Expression
lim f(x)4
lim g(x)−3
lim⁡x→5 [2f(x)−3g(x)]{\displaystyle \lim\limits_{x \to 5} \,[2f(x) - 3g(x)]}
lim f
4
lim g
−3
Answer
17{17}
Every law applied cleanly, so the answer is just arithmetic.
=lim⁡x→52f(x)−lim⁡x→53g(x) = {\displaystyle \lim\limits_{x \to 5} 2f(x) } - {\displaystyle \lim\limits_{x \to 5} 3g(x)}
Difference law: split the limit at the minus sign.
=2lim⁡x→5f(x)−3lim⁡x→5g(x) = {\displaystyle 2 \lim\limits_{x \to 5} f(x) } - {\displaystyle 3 \lim\limits_{x \to 5} g(x)}
Constant multiple law: constants come out in front.
=2(4)−3(−3)=17 = {\displaystyle 2(4) } - {\displaystyle 3(-3) } = {\displaystyle 17}
Put in the two given limits.
Check yourself
Both lim⁡x→af(x){\lim\limits_{x \to a} f(x)} and lim⁡x→ag(x){\lim\limits_{x \to a} g(x)} are 0. What does the quotient law say about lim⁡x→af(x)g(x){\lim\limits_{x \to a} \dfrac{f(x)}{g(x)}}?

Stuck on 0/0 or the squeeze theorem?

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