Exercise · ZW-LIMIT-LAWS-010 · ★★☆ Intermediate
Can the quotient law be applied to \(\lim_{x\to 3}\frac{x+1}{x-3}\)?
Choose one option.
Show answer
- A. No, because the limit of the denominator is 0
Explanation
The denominator tends to 0 and the numerator tends to 4, so the quotient law does not apply and the quotient has no finite limit.
Review the concept: Limit laws
If \(\lim_{x\to a} f(x)\) and \(\lim_{x\to a} g(x)\) exist, the limit of their sum, difference, product or quotient (with nonzero denominator) is the same operation applied to the limits.
\[\lim_{x\to a}[f(x)\,g(x)]=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)\]
More practice problems: limit laws
- ZW-LIMIT-LAWS-001 Compute lim _x→ 2 (x^2 + 3 x - 1).
- ZW-LIMIT-LAWS-002 Compute lim _x→ 4 (x - 3) (x + 1).
- ZW-LIMIT-LAWS-003 Compute lim _x→ 3 (x^2 + 1)/(x - 1).
- ZW-LIMIT-LAWS-004 Compute lim _x→ 4 √x (x - 1).
- ZW-LIMIT-LAWS-005 By the product law, what does lim _x→ a[f(x) g(x)] equal if both limits exist?
- ZW-LIMIT-LAWS-006 Besides the existence of the limits of numerator and denominator, what does the quotient law requir…
- ZW-LIMIT-LAWS-007 Compute lim _x→ 2 (x^2-4)/(x-2).
- ZW-LIMIT-LAWS-008 Compute lim _x→ 1[(x^2-1)/(x-1)+(2)/(x+1)].