Worksheet · Calculus I · Unit 1

Evaluating Limits Worksheet with Complete Solutions

Twenty-four limits that progress from substitution to algebraic, one-sided, and infinite behavior.

24 problems55 min estimated timeFoundational to intermediate progression

What is included

Practice evaluating limits with a printable student worksheet, answer key, and accessible worked solutions.

Skills assessed

  • direct substitution
  • factoring
  • rationalization
  • one-sided limits
  • limits at infinity

Prerequisites

  • function notation
  • algebraic factoring
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  1. Evaluate lim⁡x→2(3x2−5x+4)\lim_{x\to2}(3x^2-5x+4).
  2. Evaluate lim⁡x→−2(x3+4x)\lim_{x\to-2}(x^3+4x).
  3. Evaluate lim⁡t→3t2+1t+2\lim_{t\to3}\frac{t^2+1}{t+2}.
  4. Evaluate lim⁡h→0(7−4h+h2)\lim_{h\to0}(7-4h+h^2).
  5. Evaluate lim⁡u→1u+8\lim_{u\to1}\sqrt{u+8}.
  6. Evaluate lim⁡x→4x+5x\lim_{x\to4}\frac{x+5}{\sqrt{x}}.
  7. Evaluate lim⁡x→3x2−9x−3\lim_{x\to3}\frac{x^2-9}{x-3}.
  8. Evaluate lim⁡x→−4x2−16x+4\lim_{x\to-4}\frac{x^2-16}{x+4}.
  9. Evaluate lim⁡x→2x3−8x−2\lim_{x\to2}\frac{x^3-8}{x-2}.
  10. Evaluate lim⁡x→1x2+x−2x−1\lim_{x\to1}\frac{x^2+x-2}{x-1}.
  11. Evaluate lim⁡y→5y2−7y+10y−5\lim_{y\to5}\frac{y^2-7y+10}{y-5}.
  12. Evaluate lim⁡z→−1z3+1z+1\lim_{z\to-1}\frac{z^3+1}{z+1}.
  13. Evaluate lim⁡x→9x−3x−9\lim_{x\to9}\frac{\sqrt{x}-3}{x-9}.
  14. Evaluate lim⁡x→01+x−1x\lim_{x\to0}\frac{\sqrt{1+x}-1}{x}.
  15. Evaluate lim⁡t→4t−4t−2\lim_{t\to4}\frac{t-4}{\sqrt{t}-2}.
  16. Evaluate lim⁡u→0u9+u−3\lim_{u\to0}\frac{u}{\sqrt{9+u}-3}.
  17. For f(x)={x+2,x<14−x,x≥1f(x)=\begin{cases}x+2,&x<1\\4-x,&x\ge1\end{cases}, evaluate the limit as x→1−x\to1^-.
  18. For f(x)={2x,x<0x2+1,x≥0f(x)=\begin{cases}2x,&x<0\\x^2+1,&x\ge0\end{cases}, evaluate the limit as x→0+x\to0^+.
  19. For g(x)=∣x∣xg(x)=\frac{|x|}{x}, evaluate the limit as x→0−x\to0^-.
  20. For h(x)={x2,x≤26−x,x>2h(x)=\begin{cases}x^2,&x\le2\\6-x,&x>2\end{cases}, evaluate the limit as x→2x\to2.
  21. Evaluate lim⁡x→∞5x2−12x2+3\lim_{x\to\infty}\frac{5x^2-1}{2x^2+3}.
  22. Evaluate lim⁡x→∞3x+7x2+1\lim_{x\to\infty}\frac{3x+7}{x^2+1}.
  23. Evaluate lim⁡x→−∞4x3+x2x3−5\lim_{x\to-\infty}\frac{4x^3+x}{2x^3-5}.
  24. Evaluate lim⁡x→2+1x−2\lim_{x\to2^+}\frac{1}{x-2}.

Complete worked solutions

Every problem has a source-matched answer and independently reviewed derivation.

01

Problem 1: Evaluate lim⁡x→2(3x2−5x+4)\lim_{x\to2}(3x^2-5x+4).

Answer: 66

Why this method: Direct substitution matches the mathematical structure before any algebraic cleanup.

  1. The expression is continuous at the target input 22.
  2. Substitute the target and simplify to obtain 66.
02

Problem 2: Evaluate lim⁡x→−2(x3+4x)\lim_{x\to-2}(x^3+4x).

Answer: −16-16

Why this method: Direct substitution matches the mathematical structure before any algebraic cleanup.

  1. The expression is continuous at the target input −2-2.
  2. Substitute the target and simplify to obtain −16-16.
03

Problem 3: Evaluate lim⁡t→3t2+1t+2\lim_{t\to3}\frac{t^2+1}{t+2}.

Answer: 22

Why this method: Direct substitution matches the mathematical structure before any algebraic cleanup.

  1. The expression is continuous at the target input 33.
  2. Substitute the target and simplify to obtain 22.
04

Problem 4: Evaluate lim⁡h→0(7−4h+h2)\lim_{h\to0}(7-4h+h^2).

Answer: 77

Why this method: Direct substitution matches the mathematical structure before any algebraic cleanup.

  1. The expression is continuous at the target input 00.
  2. Substitute the target and simplify to obtain 77.
05

Problem 5: Evaluate lim⁡u→1u+8\lim_{u\to1}\sqrt{u+8}.

Answer: 33

Why this method: Direct substitution matches the mathematical structure before any algebraic cleanup.

  1. The expression is continuous at the target input 11.
  2. Substitute the target and simplify to obtain 33.
06

Problem 6: Evaluate lim⁡x→4x+5x\lim_{x\to4}\frac{x+5}{\sqrt{x}}.

Answer: 92\frac92

Why this method: Direct substitution matches the mathematical structure before any algebraic cleanup.

  1. The expression is continuous at the target input 44.
  2. Substitute the target and simplify to obtain 92\frac92.
07

Problem 7: Evaluate lim⁡x→3x2−9x−3\lim_{x\to3}\frac{x^2-9}{x-3}.

Answer: 66

Why this method: Factoring and cancellation matches the mathematical structure before any algebraic cleanup.

  1. Factor the vanishing numerator as (x−3)(x+3)(x-3)(x+3).
  2. Cancel the common factor only for nearby inputs, then substitute the target.
  3. The simplified expression approaches 66.
08

Problem 8: Evaluate lim⁡x→−4x2−16x+4\lim_{x\to-4}\frac{x^2-16}{x+4}.

Answer: −8-8

Why this method: Factoring and cancellation matches the mathematical structure before any algebraic cleanup.

  1. Factor the vanishing numerator as (x+4)(x−4)(x+4)(x-4).
  2. Cancel the common factor only for nearby inputs, then substitute the target.
  3. The simplified expression approaches −8-8.
09

Problem 9: Evaluate lim⁡x→2x3−8x−2\lim_{x\to2}\frac{x^3-8}{x-2}.

Answer: 1212

Why this method: Factoring and cancellation matches the mathematical structure before any algebraic cleanup.

  1. Factor the vanishing numerator as (x−2)(x2+2x+4)(x-2)(x^2+2x+4).
  2. Cancel the common factor only for nearby inputs, then substitute the target.
  3. The simplified expression approaches 1212.
10

Problem 10: Evaluate lim⁡x→1x2+x−2x−1\lim_{x\to1}\frac{x^2+x-2}{x-1}.

Answer: 33

Why this method: Factoring and cancellation matches the mathematical structure before any algebraic cleanup.

  1. Factor the vanishing numerator as (x−1)(x+2)(x-1)(x+2).
  2. Cancel the common factor only for nearby inputs, then substitute the target.
  3. The simplified expression approaches 33.
11

Problem 11: Evaluate lim⁡y→5y2−7y+10y−5\lim_{y\to5}\frac{y^2-7y+10}{y-5}.

Answer: 33

Why this method: Factoring and cancellation matches the mathematical structure before any algebraic cleanup.

  1. Factor the vanishing numerator as (y−5)(y−2)(y-5)(y-2).
  2. Cancel the common factor only for nearby inputs, then substitute the target.
  3. The simplified expression approaches 33.
12

Problem 12: Evaluate lim⁡z→−1z3+1z+1\lim_{z\to-1}\frac{z^3+1}{z+1}.

Answer: 33

Why this method: Factoring and cancellation matches the mathematical structure before any algebraic cleanup.

  1. Factor the vanishing numerator as (z+1)(z2−z+1)(z+1)(z^2-z+1).
  2. Cancel the common factor only for nearby inputs, then substitute the target.
  3. The simplified expression approaches 33.
13

Problem 13: Evaluate lim⁡x→9x−3x−9\lim_{x\to9}\frac{\sqrt{x}-3}{x-9}.

Answer: 16\frac16

Why this method: Rationalization matches the mathematical structure before any algebraic cleanup.

  1. Multiply numerator and denominator by the conjugate x+3\sqrt{x}+3.
  2. Use the difference-of-squares identity and cancel the factor that tends to zero.
  3. Substitution in the simplified expression gives 16\frac16.
14

Problem 14: Evaluate lim⁡x→01+x−1x\lim_{x\to0}\frac{\sqrt{1+x}-1}{x}.

Answer: 12\frac12

Why this method: Rationalization matches the mathematical structure before any algebraic cleanup.

  1. Multiply numerator and denominator by the conjugate 1+x+1\sqrt{1+x}+1.
  2. Use the difference-of-squares identity and cancel the factor that tends to zero.
  3. Substitution in the simplified expression gives 12\frac12.
15

Problem 15: Evaluate lim⁡t→4t−4t−2\lim_{t\to4}\frac{t-4}{\sqrt{t}-2}.

Answer: 44

Why this method: Rationalization matches the mathematical structure before any algebraic cleanup.

  1. Multiply numerator and denominator by the conjugate t+2\sqrt{t}+2.
  2. Use the difference-of-squares identity and cancel the factor that tends to zero.
  3. Substitution in the simplified expression gives 44.
16

Problem 16: Evaluate lim⁡u→0u9+u−3\lim_{u\to0}\frac{u}{\sqrt{9+u}-3}.

Answer: 66

Why this method: Rationalization matches the mathematical structure before any algebraic cleanup.

  1. Multiply numerator and denominator by the conjugate 9+u+3\sqrt{9+u}+3.
  2. Use the difference-of-squares identity and cancel the factor that tends to zero.
  3. Substitution in the simplified expression gives 66.
17

Problem 17: For f(x)={x+2,x<14−x,x≥1f(x)=\begin{cases}x+2,&x<1\\4-x,&x\ge1\end{cases}, evaluate the limit as x→1−x\to1^-.

Answer: 33

Why this method: One-sided branch analysis matches the mathematical structure before any algebraic cleanup.

  1. Approaching from the left selects the branch f(x)=x+2f(x)=x+2.
  2. As x→1−x\to1^-, x+2→3x+2\to3; the value assigned by the other branch at the endpoint does not control this one-sided limit.
  3. Therefore the result is 33.
18

Problem 18: For f(x)={2x,x<0x2+1,x≥0f(x)=\begin{cases}2x,&x<0\\x^2+1,&x\ge0\end{cases}, evaluate the limit as x→0+x\to0^+.

Answer: 11

Why this method: One-sided branch analysis matches the mathematical structure before any algebraic cleanup.

  1. Approaching from the right selects the branch f(x)=x2+1f(x)=x^2+1.
  2. As x→0+x\to0^+, x2+1→1x^2+1\to1.
  3. Therefore the result is 11.
19

Problem 19: For g(x)=∣x∣xg(x)=\frac{|x|}{x}, evaluate the limit as x→0−x\to0^-.

Answer: −1-1

Why this method: One-sided branch analysis matches the mathematical structure before any algebraic cleanup.

  1. For x<0x<0, ∣x∣=−x|x|=-x, so g(x)=(−x)/x=−1g(x)=(-x)/x=-1.
  2. Therefore the values remain −1-1 as x→0−x\to0^-.
20

Problem 20: For h(x)={x2,x≤26−x,x>2h(x)=\begin{cases}x^2,&x\le2\\6-x,&x>2\end{cases}, evaluate the limit as x→2x\to2.

Answer: 44

Why this method: Two-sided branch comparison matches the mathematical structure before any algebraic cleanup.

  1. As x→2−x\to2^-, the branch x2x^2 approaches 44.
  2. As x→2+x\to2^+, the branch 6−x6-x approaches 44.
  3. Because both one-sided limits equal 44, the two-sided limit is 44.
21

Problem 21: Evaluate lim⁡x→∞5x2−12x2+3\lim_{x\to\infty}\frac{5x^2-1}{2x^2+3}.

Answer: 52\frac52

Why this method: Dominant-term or sign analysis matches the mathematical structure before any algebraic cleanup.

  1. Identify the dominant behavior: equal degrees.
  2. Divide by the highest relevant power or use the sign of the vanishing factor.
  3. The limit is 52\frac52.
22

Problem 22: Evaluate lim⁡x→∞3x+7x2+1\lim_{x\to\infty}\frac{3x+7}{x^2+1}.

Answer: 00

Why this method: Dominant-term or sign analysis matches the mathematical structure before any algebraic cleanup.

  1. Identify the dominant behavior: denominator degree is larger.
  2. Divide by the highest relevant power or use the sign of the vanishing factor.
  3. The limit is 00.
23

Problem 23: Evaluate lim⁡x→−∞4x3+x2x3−5\lim_{x\to-\infty}\frac{4x^3+x}{2x^3-5}.

Answer: 22

Why this method: Dominant-term or sign analysis matches the mathematical structure before any algebraic cleanup.

  1. Identify the dominant behavior: equal degrees.
  2. Divide by the highest relevant power or use the sign of the vanishing factor.
  3. The limit is 22.
24

Problem 24: Evaluate lim⁡x→2+1x−2\lim_{x\to2^+}\frac{1}{x-2}.

Answer: +∞+\infty

Why this method: Dominant-term or sign analysis matches the mathematical structure before any algebraic cleanup.

  1. Identify the dominant behavior: positive denominator approaching zero.
  2. Divide by the highest relevant power or use the sign of the vanishing factor.
  3. The limit is +∞+\infty.

Common errors

  • Substituting before checking whether the form is determinate.
  • Cancelling terms instead of common factors.
  • Ignoring approach direction at jumps.