1.4 Limit Laws and Direct Substitution
Notes
If $\displaystyle \lim_{x \to a} f(x) = L$ and $\displaystyle \lim_{x \to a} g(x) = M$, then limits behave nicely with arithmetic.
- Sum and difference: $\displaystyle \lim (f \pm g) = L \pm M$.
- Constant multiple: $\displaystyle \lim c\,f(x) = cL$.
- Product: $\displaystyle \lim f(x)g(x) = LM$.
- Quotient: $\displaystyle \lim \frac{f(x)}{g(x)} = \frac{L}{M}$ if $M \neq 0$.
- Direct substitution: for polynomials and many other functions, $\displaystyle \lim_{x \to a} f(x) = f(a)$. Example: $\displaystyle \lim_{x \to -1} (x^2 - 3x + 2) = 1 + 3 + 2 = 6$.
Practice
If $\displaystyle \lim_{x \to 5} f(x) = 3$ and $\displaystyle \lim_{x \to 5} g(x) = -2$, find $\displaystyle \lim_{x \to 5} \left(2f(x) - 5g(x)\right)$.
$2(3) - 5(-2) = 6 + 10 = 16$.
Find $\displaystyle \lim_{x \to -1} (x^2 - 3x + 2)$.
Substitute $x = -1$: $1 + 3 + 2 = 6$.
Find $\displaystyle \lim_{x \to 2} \frac{x^2 + 1}{x + 3}$.
The denominator is not zero at 2, so substitute: $\displaystyle \frac{4 + 1}{2 + 3} = 1$.
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