[Calculus I/Limits #2] Mastering the 4 Indeterminate Forms: 15 Essential Practice Problems & Twin Worksheet
π [Calculus I] Limits of Functions: 15 Essential Practice Problems on the 4 Indeterminate Forms
Mastering limits of functions goes beyond mechanical algebraic manipulations—it starts with understanding the structural behavior of the 4 Indeterminate Forms ($\frac{0}{0}, \frac{\infty}{\infty}, \infty-\infty, 0 \times \infty$) and controlling zero factors. This curated problem set is structured across 4 progressive tiers, from double rationalization and negative infinity substitutions to advanced polynomial degree inferences and continuous extensions.
π Tier Breakdown
- STEP 1. Core Foundations (Q1~Q4): Double radical rationalization, multi-term $\infty-\infty$, common denominators, and variable substitution.
- STEP 2. Application & Variation (Q5~Q8): Negative infinity ($x \to -\infty$), one-sided limits with absolute values, Squeeze Theorem with floor functions, and polynomial coefficient determination.
- STEP 3. Advanced Reasoning (Q9~Q12): Polynomial identity analysis, geometric locus limits, and derivative-limit continuous definitions.
- STEP 4. CSAT Advanced Killers (Q13~Q15): Polynomial degree and $n$-th order root deductions, symmetric derivative continuity, and new functions defined by limits.
※ Solve each problem independently before clicking [View Answer & Solution] to review the step-by-step solution!
STEP 1. Core Foundations (Q1~Q4)
[Problem 01] Evaluate the following limit:
π View Answer & Solution
Solution: Rationalize both numerator and denominator simultaneously.
$$\begin{aligned} &\lim_{x \to 0} \frac{\{(1+x+x^2) - (1-x+x^2)\}(\sqrt{4+x} + \sqrt{4-x})}{\{(4+x) - (4-x)\}(\sqrt{1+x+x^2} + \sqrt{1-x+x^2})} \\ &= \lim_{x \to 0} \frac{2x \cdot (\sqrt{4+x} + \sqrt{4-x})}{2x \cdot (\sqrt{1+x+x^2} + \sqrt{1-x+x^2})} = \frac{2+2}{1+1} \times \frac{2}{1} = 4 \end{aligned}$$
[Problem 02] Evaluate the following limit:
π View Answer & Solution
Solution: Group $3x$ as $x + 2x$ and rationalize each part separately.
$$\lim_{x \to \infty} (\sqrt{x^2+4x+1}-x) + \lim_{x \to \infty} (\sqrt{4x^2+8x-3}-2x) = \frac{4}{1+1} + \frac{8}{2+2} = 2 + 2 = 4$$
[Problem 03] Given that $\lim_{x \to 0} \frac{1}{x} \left( \frac{1}{\sqrt{1+ax}} - 1 \right) = -2$, find the value of constant $a$.
π View Answer & Solution
Solution: Find a common denominator and rationalize the numerator.
$$\frac{1-\sqrt{1+ax}}{\sqrt{1+ax}} = \frac{-ax}{\sqrt{1+ax}(1+\sqrt{1+ax})}$$ $$\lim_{x \to 0} \frac{1}{x} \cdot \frac{-ax}{\sqrt{1+ax}(1+\sqrt{1+ax})} = \frac{-a}{1 \cdot 2} = -\frac{a}{2} = -2 \implies a = 4$$
[Problem 04] For positive real numbers $x$, evaluate the following limit:
π View Answer & Solution
Solution: Rationalize the expression inside the parenthesis.
$$\lim_{x \to \infty} \frac{4x}{\sqrt{x^2+2} + \sqrt{x^2-2}} = \lim_{x \to \infty} \frac{4}{\sqrt{1+\frac{2}{x^2}} + \sqrt{1-\frac{2}{x^2}}} = \frac{4}{1+1} = 2$$
STEP 2. Application & Variation (Q5~Q8)
[Problem 05] Given that $\lim_{x \to -\infty} \left( \sqrt{4x^2 + ax + 1} + 2x \right) = 3$, find the value of constant $a$.
π View Answer & Solution
Solution: Substitute $x = -t$ ($t \to \infty$) and rationalize.
$$\lim_{t \to \infty} (\sqrt{4t^2-at+1} - 2t) = \lim_{t \to \infty} \frac{-at+1}{\sqrt{4t^2-at+1}+2t} = \frac{-a}{2+2} = -\frac{a}{4} = 3 \implies a = -12$$
[Problem 06] For the function $f(x) = \frac{|x^2 - x - 2|}{x^2 + ax + b}$, let $\lim_{x \to 2^+} f(x) = \alpha$ and $\lim_{x \to 2^-} f(x) = \beta$. If $\alpha$ and $\beta$ are non-zero real numbers such that $\alpha + \beta = 0$ and $\alpha = 1$, find the value of $a+b$.
π View Answer & Solution
Solution: Since $x^2-x-2 = (x-2)(x+1)$, for the limit to exist at $x \to 2$, the denominator must have $(x-2)$ as a factor.
Let denominator $= (x-2)(x-k)$. Then $\alpha = \lim_{x \to 2^+} \frac{(x-2)(x+1)}{(x-2)(x-k)} = \frac{3}{2-k} = 1 \implies k = -1$.
Thus, the denominator is $(x-2)(x+1) = x^2 - x - 2 \implies a = -1, b = -2 \implies a+b = -3$.
[Problem 07] For a real number $x$, let $[x]$ denote the greatest integer less than or equal to $x$. Evaluate the following limit:
π View Answer & Solution
Solution: As $x \to \infty$, $\frac{6x+1}{x} = 6 + \frac{1}{x} \to 6^+ \implies [6+\frac{1}{x}] = 6$, and $\frac{6x-2}{x} = 6 - \frac{2}{x} \to 6^- \implies [6-\frac{2}{x}] = 5$.
Thus, the bracketed term converges to $6 - 5 = 1$.
$$\lim_{x \to \infty} \frac{x}{3x+1} \cdot 1 = \frac{1}{3}$$
[Problem 08] A polynomial function $f(x)$ satisfies the following two conditions. Find the value of $f(2)$.
- • (A) $\lim_{x \to \infty} \frac{f(x) - 2x^3}{x^2 + 1} = 4$
- • (B) $\lim_{x \to 0} \frac{f(x)}{x} = -6$
π View Answer & Solution
Solution:
1. From (A), $f(x) = 2x^3 + 4x^2 + ax + b$.
2. From (B), $f(0) = 0 \implies b = 0$, and $\lim_{x \to 0} \frac{f(x)}{x} = a = -6$.
3. $f(x) = 2x^3 + 4x^2 - 6x \implies f(2) = 16 + 16 - 12 = 20$.
STEP 3. Advanced Reasoning (Q9~Q12)
[Problem 09] For a cubic function $f(x)$ with leading coefficient $1$, define $g(x) = \frac{f(x)}{x-1}$ ($x \neq 1$) satisfying the following conditions:
- • (A) $\lim_{x \to 1} g(x) = 6$
- • (B) The sum of the two distinct real roots of $g(x) = 6$ is $3$.
Find the value of $f(5)$.
π View Answer & Solution
Solution: Let $g(x) = x^2 + ax + b$.
1. From (A), $g(1) = 1 + a + b = 6 \implies b = 5 - a$.
2. From (B), $x^2 + ax + (5-a) = 6 \implies x^2 + ax - (a+1) = 0$. By Vieta's formulas, sum of roots is $-a = 3 \implies a = -3, b = 8$.
3. $f(x) = (x-1)(x^2 - 3x + 8) \implies f(5) = 4 \times (25 - 15 + 8) = 4 \times 18 = 72$.
[Problem 10] For a positive real number $t$, let the line $y = tx + 1$ intersect the parabola $y = x^2$ at two points $A$ and $B$. Let $L(t)$ be the length of segment $AB$. Evaluate the following limit:
π View Answer & Solution
Solution: Let the roots of $x^2 - tx - 1 = 0$ be $\alpha, \beta$. Then $(\beta-\alpha)^2 = t^2+4$.
$L(t) = \sqrt{1+t^2} \cdot (\beta-\alpha) = \sqrt{(t^2+1)(t^2+4)} = \sqrt{t^4+5t^2+4}$.
$$\lim_{t \to \infty} (\sqrt{t^4+5t^2+4} - t^2) = \lim_{t \to \infty} \frac{5t^2+4}{\sqrt{t^4+5t^2+4}+t^2} = \frac{5}{1+1} = \frac{5}{2}$$
[Problem 11] For a quadratic function $f(x)$ with leading coefficient $1$, given that $\lim_{x \to 2} \frac{f(x)}{x-2} = 4$, evaluate the following limit:
π View Answer & Solution
Solution: $f(2) = 0$ and $\lim_{x \to 2} \frac{f(x)}{x-2} = 4$.
$$\lim_{x \to 2} \frac{\{f(x)\}^2}{(x-2)(x+2)} = \lim_{x \to 2} \left( \frac{f(x)}{x-2} \cdot \frac{f(x)}{x+2} \right) = 4 \cdot \frac{f(2)}{4} = 4 \cdot \frac{0}{4} = 0$$
[Problem 12] On the coordinate plane, let $P$ be a point in the first quadrant on the circle $x^2 + y^2 = r^2$ ($r>0$). The tangent line at $P$ intersects the $x$-axis at point $Q$, and $H$ is the projection of $P$ onto the $x$-axis. As point $P$ approaches $(r, 0)$ along the circle, find the value of $\lim_{P \to (r,0)} \frac{\overline{HQ}}{\overline{HP}^2}$.
π View Answer & Solution
Solution: Let $P(t, \sqrt{r^2-t^2})$ ($t \to r^-$). The $x$-intercept of the tangent line is $Q\left(\frac{r^2}{t}, 0\right)$.
$\overline{HQ} = \frac{r^2}{t} - t = \frac{r^2-t^2}{t}$, $\overline{HP}^2 = r^2-t^2$.
$$\lim_{t \to r^-} \frac{\frac{r^2-t^2}{t}}{r^2-t^2} = \lim_{t \to r^-} \frac{1}{t} = \frac{1}{r}$$
STEP 4. CSAT Advanced Killers (Q13~Q15)
[Problem 13] A polynomial function $f(x)$ satisfies the following two conditions:
- • (A) $\lim_{x \to \infty} \frac{f(x) - x^4}{x^3 + 2x} = 2$
- • (B) For a natural number $n$, the maximum value of $n$ satisfying $\lim_{x \to 0} \frac{f(x)}{x^n} = L$ ($L \neq 0$) is $2$, and $L = -8$.
Find the sum of all real roots of the equation $f'(x) = 0$.
π View Answer & Solution
Solution:
1. From (A), $f(x) = x^4 + 2x^3 + ax^2 + bx + c$.
2. From (B), the lowest degree term is $-8x^2 \implies b = c = 0, a = -8 \implies f(x) = x^4 + 2x^3 - 8x^2$.
3. $f'(x) = 4x^3 + 6x^2 - 16x = 2x(2x^2 + 3x - 8) = 0$.
4. Since $2x^2 + 3x - 8 = 0$ has distinct real roots, the sum of all real roots is $0 + \left(-\frac{3}{2}\right) = -\frac{3}{2}$.
[Problem 14] For a cubic function $f(x)$ with leading coefficient $1$, define the function $g(x)$ as follows:
Given that $g(x)$ is continuous on the set of all real numbers and satisfies the following conditions, find the value of $f(4)$:
- • (A) $\lim_{x \to 1} \frac{g(x)}{x-1} = 0$
- • (B) $\lim_{x \to \infty} \frac{g(x)}{3x^2 - 2x} = 1$
π View Answer & Solution
Solution:
1. $g(x)$ represents the symmetric derivative of $|f(x)|$. For $g(x)$ to be continuous everywhere, $f(x)$ must have no sharp non-differentiable crossing points, requiring a triple root form: $f(x) = (x-\alpha)^3$.
2. Then $g(x) = 3(x-\alpha)^2$, which satisfies condition (B) with leading coefficient ratio $1$.
3. From (A), $g(1) = 0 \implies \alpha = 1 \implies f(x) = (x-1)^3$.
4. $f(4) = (4-1)^3 = 27$.
[Problem 15] For a real number $t$, a quadratic function $f(x)$ with leading coefficient $1$ satisfies $\lim_{x \to t} \frac{f(x) - f(t)}{|x-t|} = h(t)$. If $h(t)$ has a limit only at $t = 2$ and the minimum value of $f(x)$ is $-4$, evaluate the following limit:
π View Answer & Solution
Solution:
1. The right-hand limit is $f'(t)$ and the left-hand limit is $-f'(t)$. For the limit to exist, $f'(t) = -f'(t) \implies f'(t) = 0$.
2. Since this holds only at $t=2$, the vertex axis of symmetry is $x=2$. With minimum value $-4$, $f(x) = (x-2)^2 - 4 = x^2 - 4x$.
3. $\lim_{x \to 5} \frac{x^2-4x}{x-4} = \frac{25-20}{5-4} = 5$.
π [Self-Test] Twin Practice Worksheet (15 Variant Problems)
Now that you've reviewed the core solutions, test your true mastery with these 15 variant problems featuring identical logical structures and difficulty levels with modified numerical values!
[Variant] STEP 1. Core Foundations (Q01~Q04)
[Variant 01] Evaluate the following limit:
π View Answer & Solution
Solution: $\lim_{x \to 0} \frac{4x \cdot (\sqrt{9+x}+\sqrt{9-x})}{2x \cdot (\sqrt{1+2x+x^2}+\sqrt{1-2x+x^2})} = \frac{4 \times (3+3)}{2 \times (1+1)} = \frac{24}{4} = 12$.
[Variant 02] Evaluate the following limit:
π View Answer & Solution
Solution: Group as $(\sqrt{x^2+6x+2}-x) + (\sqrt{9x^2+12x-1}-3x) \implies \frac{6}{1+1} + \frac{12}{3+3} = 3 + 2 = 5$.
[Variant 03] Given that $\lim_{x \to 0} \frac{1}{x} \left( \frac{1}{\sqrt{1+kx}} - 1 \right) = -3$, find the value of constant $k$.
π View Answer & Solution
Solution: $\lim_{x \to 0} \frac{-k}{\sqrt{1+kx}(1+\sqrt{1+kx})} = -\frac{k}{2} = -3 \implies k = 6$.
[Variant 04] For positive real numbers $x$, evaluate the following limit:
π View Answer & Solution
Solution: $\lim_{x \to \infty} \frac{12x}{\sqrt{x^2+6}+\sqrt{x^2-6}} = \frac{12}{1+1} = 6$.
[Variant] STEP 2. Application & Variation (Q05~Q08)
[Variant 05] Given that $\lim_{x \to -\infty} \left( \sqrt{9x^2 + ax + 2} + 3x \right) = 2$, find the value of constant $a$.
π View Answer & Solution
Solution: Substitute $x = -t$ and rationalize $\implies \lim_{t \to \infty} \frac{-at+2}{\sqrt{9t^2-at+2}+3t} = \frac{-a}{3+3} = -\frac{a}{6} = 2 \implies a = -12$.
[Variant 06] For $f(x) = \frac{|x^2 - 4|}{x^2 + ax + b}$, given that $\lim_{x \to 2^+} f(x) = 2$ and $\lim_{x \to 2^-} f(x) = -2$, find the value of $a+b$.
π View Answer & Solution
Solution: Let denominator be $(x-2)(x-k)$. Then $\frac{4}{2-k} = 2 \implies k = 0$.
Denominator $= x(x-2) = x^2 - 2x \implies a = -2, b = 0 \implies a+b = -2$.
[Variant 07] Evaluate the following limit:
π View Answer & Solution
Solution: $[4 + \frac{1}{x}] = 4$, $[4 - \frac{1}{x}] = 3 \implies 4 - 3 = 1$.
$$\lim_{x \to \infty} \frac{x}{2x+3} \cdot 1 = \frac{1}{2}$$
[Variant 08] A polynomial $f(x)$ satisfies $\lim_{x \to \infty} \frac{f(x) - 3x^3}{x^2+1} = 2$ and $\lim_{x \to 0} \frac{f(x)}{x} = -4$. Find the value of $f(1)$.
π View Answer & Solution
Solution: $f(x) = 3x^3 + 2x^2 - 4x \implies f(1) = 3 + 2 - 4 = 1$.
[Variant] STEP 3. Advanced Reasoning (Q09~Q12)
[Variant 09] For a cubic function $f(x)$ with leading coefficient $1$, let $g(x) = \frac{f(x)}{x-2}$ ($x \neq 2$). Given that $\lim_{x \to 2} g(x) = 8$ and the sum of the two real roots of $g(x) = 8$ is $5$, find the value of $f(4)$.
π View Answer & Solution
Solution: $g(x) = x^2 - 5x + 14 \implies f(x) = (x-2)(x^2 - 5x + 14) \implies f(4) = 2 \times (16 - 20 + 14) = 20$.
[Variant 10] Let $L(t)$ be the distance between the two intersection points of $y = tx + 2$ and $y = x^2$. Evaluate the following limit:
π View Answer & Solution
Solution: $L(t) = \sqrt{(t^2+1)(t^2+8)} = \sqrt{t^4+9t^2+8} \implies \lim_{t \to \infty} \frac{9t^2+8}{\sqrt{t^4+9t^2+8}+t^2} = \frac{9}{2}$.
[Variant 11] For a quadratic function $f(x)$ with leading coefficient $1$, given that $\lim_{x \to 3} \frac{f(x)}{x-3} = 6$, evaluate the limit $\lim_{x \to 3} \frac{\{f(x)\}^2}{x^2 - 9}$.
π View Answer & Solution
Solution: $\lim_{x \to 3} \left( \frac{f(x)}{x-3} \cdot \frac{f(x)}{x+3} \right) = 6 \cdot \frac{f(3)}{6} = 6 \cdot 0 = 0$.
[Variant 12] Let $Q$ be the $x$-intercept of the tangent line at $P(t, \sqrt{4-t^2})$ on the circle $x^2 + y^2 = 4$, and let $H$ be the projection of $P$ onto the $x$-axis. Evaluate $\lim_{t \to 2^-} \frac{\overline{HQ}}{\overline{HP}^2}$.
π View Answer & Solution
Solution: $\overline{HQ} = \frac{4-t^2}{t}, \overline{HP}^2 = 4-t^2 \implies \lim_{t \to 2^-} \frac{1}{t} = \frac{1}{2}$.
[Variant] STEP 4. CSAT Advanced Killers (Q13~Q15)
[Variant 13] A polynomial $f(x)$ satisfies $\lim_{x \to \infty} \frac{f(x) - x^4}{x^3 + 1} = 4$ and $\lim_{x \to 0} \frac{f(x)}{x^2} = -6$. Find the sum of all real roots of $f'(x) = 0$.
π View Answer & Solution
Solution: $f(x) = x^4 + 4x^3 - 6x^2 \implies f'(x) = 4x^3 + 12x^2 - 12x = 4x(x^2 + 3x - 3) = 0$.
Sum of real roots is $0 + (-3) = -3$.
[Variant 14] For a cubic function $f(x)$ with leading coefficient $1$, let $g(x) = \lim_{t \to 0^+} \frac{|f(x+t)| - |f(x-t)|}{2t}$ be continuous everywhere. If $\lim_{x \to 2} \frac{g(x)}{x-2} = 0$ and $\lim_{x \to \infty} \frac{g(x)}{3x^2} = 1$, find the value of $f(5)$.
π View Answer & Solution
Solution: $f(x) = (x-2)^3 \implies f(5) = (5-2)^3 = 27$.
[Variant 15] A quadratic function $f(x)$ with leading coefficient $1$ satisfies $\lim_{x \to t} \frac{f(x) - f(t)}{|x-t|} = h(t)$. If $h(t)$ has a limit only at $t = 3$ and the minimum value of $f(x)$ is $-9$, evaluate $\lim_{x \to 7} \frac{f(x)}{x-6}$.
π View Answer & Solution
Solution: $f(x) = (x-3)^2 - 9 = x^2 - 6x \implies \lim_{x \to 7} \frac{x(x-6)}{x-6} = 7$.
✍️ Final Insight from Yul
In calculus, the 4 Indeterminate Forms should never be tackled by merely memorizing algebraic manipulation tricks.
True mathematical proficiency begins when you can read the convergence speed of zero factors and identify which highest-degree term dominates the behavior at infinity. Once this foundational vision is solid, advanced polynomial deduction and derivative continuity naturally fall into place.
Work through each algebraic step with discipline. When the structural elegance of mathematical expressions becomes visible, your mathematical intuition and confidence will reach an elite standard.
π [Calculus I] Mastering Limits of Functions Series
- Part 1: [Core Concepts] The 4 Indeterminate Forms & 80% Error Trap Breakdown
- Part 2: [Practice & Test] 15 Essential Problems + Twin Self-Test Worksheet (Current Post)
Comments
Post a Comment