AP Calculus BC Flashcards: Determining Limits Using Algebraic Manipulation

What this deck covers

This deck focuses on Determining Limits Using Algebraic Manipulation, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is \lim_{x\to 1}\frac{x-1}{\sqrt{x}-1}?

Answer: $2$. Factor numerator: $\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1} = \sqrt{x}+1$.

Flashcard 2: What is \lim_{x\to 0}\frac{\frac{1}{x+1}-1}{x}?

Answer: $-1$. Simplify to $\frac{-x}{x(x+1)}=\frac{-1}{x+1}$, then substitute $x=0$.

Flashcard 3: What is \lim_{x\to 0}\frac{x}{\sqrt{1+x}-1} after rationalizing?

Answer: $2$. After rationalizing, get $\frac{x(\sqrt{1+x}+1)}{x}=\sqrt{1+x}+1$, substitute $x=0$.

Flashcard 4: What is \lim_{x\to 2}\frac{x^2-4}{x^2-3x+2}?

Answer: $4$. Factor: $\frac{(x-2)(x+2)}{(x-2)(x-1)} = \frac{x+2}{x-1}$, then substitute.

Flashcard 5: What algebraic technique is most appropriate when \lim_{x\to a}\frac{f(x)}{g(x)} gives \frac{0}{0} and $f,g$ are polynomials?

Answer: Factor numerator and denominator, then cancel the common factor. This eliminates the indeterminate form $\frac{0}{0}$.

Flashcard 6: What is the standard limit value \lim_{u\to 0}\frac{1-\cos u}{u^2} used in algebraic manipulation?

Answer: $\frac{1}{2}$. Derived from $\sin^2u=\frac{1-\cos(2u)}{2}$ and double angle formula.

Flashcard 7: What is \lim_{x\to 1}\frac{\frac{1}{x}-1}{x-1}?

Answer: $-1$. Rewrite as $\frac{1-x}{x(x-1)} = \frac{-(x-1)}{x(x-1)} = \frac{-1}{x}$.

Flashcard 8: What is \lim_{x\to 0}\frac{(x+1)^5-1}{x}?

Answer: $5$. This is the derivative of $(x+1)^5$ at $x=0$ using the difference quotient.

Flashcard 9: What is \lim_{x\to 0}\frac{1-\cos(3x)}{x^2} using standard trig limit algebra?

Answer: $\frac{9}{2}$. Use $\frac{1-\cos(3x)}{x^2}=9\cdot\frac{1-\cos(3x)}{(3x)^2}$ with standard limit $\frac{1}{2}$.

Flashcard 10: What is \lim_{x\to 0}\frac{\sqrt{9+x}-3}{x}?

Answer: $\frac{1}{6}$. Rationalize to get $\frac{x}{x(\sqrt{9+x}+3)} = \frac{1}{\sqrt{9+x}+3}$.

Flashcard 11: What is \lim_{x\to -2}\frac{x^2+5x+6}{x+2}?

Answer: $1$. Factor: $\frac{(x+2)(x+3)}{x+2} = x+3$, then substitute $x=-2$.

Flashcard 12: What is \lim_{x\to 3}\frac{x^2-9}{x-3}?

Answer: $6$. Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x=3$.

Flashcard 13: What is \lim_{h\to 0}\frac{(a+h)^2-a^2}{h}?

Answer: $2a$. Expand: $\frac{a^2+2ah+h^2-a^2}{h} = \frac{2ah+h^2}{h} = 2a+h$.

Flashcard 14: What is \lim_{h\to 0}\frac{(a+h)^3-a^3}{h}?

Answer: $3a^2$. Expand and simplify: $\frac{3a^2h+3ah^2+h^3}{h} = 3a^2+3ah+h^2$.

Flashcard 15: What is \lim_{x\to 3}\frac{x^2-9}{x-3}?

Answer: $6$. Factor as $\frac{(x-3)(x+3)}{x-3}$, cancel to get $x+3$, then substitute $x=3$.

Flashcard 16: What is \lim_{x\to 2}\frac{x^2-4}{x-2}?

Answer: $4$. Factor as $\frac{(x-2)(x+2)}{x-2}$, cancel to get $x+2$, then substitute $x=2$.

Flashcard 17: What algebraic technique is most appropriate for \lim_{x\to a}\frac{\sqrt{u(x)}-\sqrt{v(x)}}{w(x)} when direct substitution gives \frac{0}{0}?

Answer: Multiply by the conjugate of the numerator to rationalize. Creates a difference of squares in the numerator.

Flashcard 18: What identity is used to factor $a^3-b^3$ for limit simplification?

Answer: $a^3-b^3=(a-b)(a^2+ab+b^2)$. This is the difference of cubes factorization formula.

Flashcard 19: What is \lim_{x\to 2}\frac{x^3-8}{x-2}?

Answer: $12$. Factor as $\frac{(x-2)(x^2+2x+4)}{x-2}$, cancel, then substitute to get $4+4+4$.

Flashcard 20: What is \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x}?

Answer: $\frac{1}{2}$. Multiply by conjugate $\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$ to get $\frac{1}{\sqrt{1+x}+1}$.

Flashcard 21: What is \lim_{x\to 2}\frac{x^3-8}{x-2}?

Answer: $12$. Use $x^3-8=(x-2)(x^2+2x+4)$, cancel, then substitute $x=2$.

Flashcard 22: What is \lim_{x\to 0}\frac{(1+x)^5-1}{x}?

Answer: $5$. This is the derivative of $(1+x)^5$ at $x=0$ using difference quotient.

Flashcard 23: What is \lim_{x\to 0}\frac{\sqrt{x+9}-3}{x}?

Answer: $\frac{1}{6}$. Multiply by conjugate $\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}$ to get $\frac{1}{\sqrt{x+9}+3}$.

Flashcard 24: What is \lim_{x\to 1}\frac{x^3-1}{x-1}?

Answer: $3$. Factor as $\frac{(x-1)(x^2+x+1)}{x-1}$, cancel, then substitute to get $1+1+1$.

Flashcard 25: What is \lim_{x\to 0}\frac{\tan(4x)}{x} using \tan u=\frac{\sin u}{\cos u}?

Answer: $4$. Rewrite as $\frac{\sin(4x)}{x\cos(4x)}=4\cdot\frac{\sin(4x)}{4x}\cdot\frac{1}{\cos(4x)}$.

Flashcard 26: What is the standard limit value \lim_{u\to 0}\frac{\sin u}{u} used for algebraic substitution?

Answer: $1$. Fundamental trigonometric limit proven using squeeze theorem.

Flashcard 27: What is \lim_{x\to -1}\frac{x^3+1}{x+1}?

Answer: $3$. Use $x^3+1=(x+1)(x^2-x+1)$, cancel, then substitute $x=-1$.

Flashcard 28: What is \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x}?

Answer: $\frac{1}{2}$. Multiply by conjugate: $\frac{x}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1}$.

Flashcard 29: What is \lim_{x\to 1}\frac{x^3-1}{x-1}?

Answer: $3$. Use $x^3-1=(x-1)(x^2+x+1)$, cancel, then substitute $x=1$.

Flashcard 30: What is \lim_{x\to -1}\frac{x^2+3x+2}{x+1}?

Answer: $1$. Factor as $\frac{(x+1)(x+2)}{x+1}$, cancel to get $x+2$, then substitute $x=-1$.

Flashcard 31: What algebraic technique is most common when \lim_{x\to a}\frac{f(x)}{g(x)} gives \frac{0}{0} and polynomials are involved?

Answer: Factor numerator and denominator, then cancel the common factor. Factoring reveals removable discontinuities when direct substitution yields $\frac{0}{0}$.

Flashcard 32: What is \lim_{x\to 0}\frac{\sin(5x)}{x} using standard trig limit algebra?

Answer: $5$. Rewrite as $5\cdot\frac{\sin(5x)}{5x}$ and use $\lim_{u\to 0}\frac{\sin u}{u}=1$.

Flashcard 33: What identity is used to factor $a^3+b^3$ for limit simplification?

Answer: $a^3+b^3=(a+b)(a^2-ab+b^2)$. This is the sum of cubes factorization formula.

Flashcard 34: What is the conjugate of $a+b$ that is used to simplify radicals in limits?

Answer: $a-b$. Multiplying by the conjugate eliminates square roots in the denominator.

Flashcard 35: What is \lim_{x\to 1}\frac{x^2-1}{x^3-1}?

Answer: $\frac{2}{3}$. Factor as $\frac{(x-1)(x+1)}{(x-1)(x^2+x+1)}$, cancel and substitute.

Flashcard 36: What is \lim_{x\to 0}\frac{\frac{1}{x+1}-1}{x}?

Answer: $-1$. Simplify: $\frac{\frac{1-(x+1)}{x+1}}{x} = \frac{-x}{x(x+1)} = \frac{-1}{x+1}$.

Flashcard 37: What is \lim_{x\to 0}\frac{(1+2x)^3-1}{x}?

Answer: $6$. Derivative of $(1+2x)^3$ at $x=0$: $3(1+2x)^2 \cdot 2 = 6$.

Flashcard 38: What is \lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}?

Answer: $\frac{1}{4}$. Multiply by conjugate to get $\frac{1}{\sqrt{x}+2}$, then substitute $x=4$.

Flashcard 39: What is \lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}?

Answer: $\frac{1}{4}$. Multiply by $\frac{\sqrt{x}+2}{\sqrt{x}+2}$ to get $\frac{1}{\sqrt{x}+2}$.