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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Evaluate $\lim_{x\to2}(3x^2-5x+4)$ .
Evaluate $\lim_{x\to-2}(x^3+4x)$ .
Evaluate $\lim_{t\to3}\frac{t^2+1}{t+2}$ .
Evaluate $\lim_{h\to0}(7-4h+h^2)$ .
Evaluate $\lim_{u\to1}\sqrt{u+8}$ .
Evaluate $\lim_{x\to4}\frac{x+5}{\sqrt{x}}$ .
Evaluate $\lim_{x\to3}\frac{x^2-9}{x-3}$ .
Evaluate $\lim_{x\to-4}\frac{x^2-16}{x+4}$ .
Evaluate $\lim_{x\to2}\frac{x^3-8}{x-2}$ .
Evaluate $\lim_{x\to1}\frac{x^2+x-2}{x-1}$ .
Evaluate $\lim_{y\to5}\frac{y^2-7y+10}{y-5}$ .
Evaluate $\lim_{z\to-1}\frac{z^3+1}{z+1}$ .
Evaluate $\lim_{x\to9}\frac{\sqrt{x}-3}{x-9}$ .
Evaluate $\lim_{x\to0}\frac{\sqrt{1+x}-1}{x}$ .
Evaluate $\lim_{t\to4}\frac{t-4}{\sqrt{t}-2}$ .
Evaluate $\lim_{u\to0}\frac{u}{\sqrt{9+u}-3}$ .
For $f(x)=\begin{cases}x+2,&x<1\\4-x,&x\ge1\end{cases}$ , evaluate the limit as $x\to1^-$ .
For $f(x)=\begin{cases}2x,&x<0\\x^2+1,&x\ge0\end{cases}$ , evaluate the limit as $x\to0^+$ .
For $g(x)=\frac{|x|}{x}$ , evaluate the limit as $x\to0^-$ .
For $h(x)=\begin{cases}x^2,&x\le2\\6-x,&x>2\end{cases}$ , evaluate the limit as $x\to2$ .
Evaluate $\lim_{x\to\infty}\frac{5x^2-1}{2x^2+3}$ .
Evaluate $\lim_{x\to\infty}\frac{3x+7}{x^2+1}$ .
Evaluate $\lim_{x\to-\infty}\frac{4x^3+x}{2x^3-5}$ .
Evaluate $\lim_{x\to2^+}\frac{1}{x-2}$ .

Complete worked solutions

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

01

Problem 1: Evaluate $\lim_{x\to2}(3x^2-5x+4)$ .

Answer: $6$

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

The expression is continuous at the target input $2$ .
Substitute the target and simplify to obtain $6$ .

02

Problem 2: Evaluate $\lim_{x\to-2}(x^3+4x)$ .

Answer: $-16$

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

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

03

Problem 3: Evaluate $\lim_{t\to3}\frac{t^2+1}{t+2}$ .

Answer: $2$

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

The expression is continuous at the target input $3$ .
Substitute the target and simplify to obtain $2$ .

04

Problem 4: Evaluate $\lim_{h\to0}(7-4h+h^2)$ .

Answer: $7$

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

The expression is continuous at the target input $0$ .
Substitute the target and simplify to obtain $7$ .

05

Problem 5: Evaluate $\lim_{u\to1}\sqrt{u+8}$ .

Answer: $3$

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

The expression is continuous at the target input $1$ .
Substitute the target and simplify to obtain $3$ .

06

Problem 6: Evaluate $\lim_{x\to4}\frac{x+5}{\sqrt{x}}$ .

Answer: $\frac92$

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

The expression is continuous at the target input $4$ .
Substitute the target and simplify to obtain $\frac92$ .

07

Problem 7: Evaluate $\lim_{x\to3}\frac{x^2-9}{x-3}$ .

Answer: $6$

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

Factor the vanishing numerator as $(x-3)(x+3)$ .
Cancel the common factor only for nearby inputs, then substitute the target.
The simplified expression approaches $6$ .

08

Problem 8: Evaluate $\lim_{x\to-4}\frac{x^2-16}{x+4}$ .

Answer: $-8$

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

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

09

Problem 9: Evaluate $\lim_{x\to2}\frac{x^3-8}{x-2}$ .

Answer: $12$

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

Factor the vanishing numerator as $(x-2)(x^2+2x+4)$ .
Cancel the common factor only for nearby inputs, then substitute the target.
The simplified expression approaches $12$ .

10

Problem 10: Evaluate $\lim_{x\to1}\frac{x^2+x-2}{x-1}$ .

Answer: $3$

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

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

11

Problem 11: Evaluate $\lim_{y\to5}\frac{y^2-7y+10}{y-5}$ .

Answer: $3$

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

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

12

Problem 12: Evaluate $\lim_{z\to-1}\frac{z^3+1}{z+1}$ .

Answer: $3$

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

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

13

Problem 13: Evaluate $\lim_{x\to9}\frac{\sqrt{x}-3}{x-9}$ .

Answer: $\frac16$

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

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

14

Problem 14: Evaluate $\lim_{x\to0}\frac{\sqrt{1+x}-1}{x}$ .

Answer: $\frac12$

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

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

15

Problem 15: Evaluate $\lim_{t\to4}\frac{t-4}{\sqrt{t}-2}$ .

Answer: $4$

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

Multiply numerator and denominator by the conjugate $\sqrt{t}+2$ .
Use the difference-of-squares identity and cancel the factor that tends to zero.
Substitution in the simplified expression gives $4$ .

16

Problem 16: Evaluate $\lim_{u\to0}\frac{u}{\sqrt{9+u}-3}$ .

Answer: $6$

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

Multiply numerator and denominator by the conjugate $\sqrt{9+u}+3$ .
Use the difference-of-squares identity and cancel the factor that tends to zero.
Substitution in the simplified expression gives $6$ .

17

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

Answer: $3$

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

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

18

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

Answer: $1$

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

Approaching from the right selects the branch $f(x)=x^2+1$ .
As $x\to0^+$ , $x^2+1\to1$ .
Therefore the result is $1$ .

19

Problem 19: For $g(x)=\frac{|x|}{x}$ , evaluate the limit as $x\to0^-$ .

Answer: $-1$

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

For $x<0$ , $|x|=-x$ , so $g(x)=(-x)/x=-1$ .
Therefore the values remain $-1$ as $x\to0^-$ .

20

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

Answer: $4$

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

As $x\to2^-$ , the branch $x^2$ approaches $4$ .
As $x\to2^+$ , the branch $6-x$ approaches $4$ .
Because both one-sided limits equal $4$ , the two-sided limit is $4$ .

21

Problem 21: Evaluate $\lim_{x\to\infty}\frac{5x^2-1}{2x^2+3}$ .

Answer: $\frac52$

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

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

22

Problem 22: Evaluate $\lim_{x\to\infty}\frac{3x+7}{x^2+1}$ .

Answer: $0$

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

Identify the dominant behavior: denominator degree is larger.
Divide by the highest relevant power or use the sign of the vanishing factor.
The limit is $0$ .

23

Problem 23: Evaluate $\lim_{x\to-\infty}\frac{4x^3+x}{2x^3-5}$ .

Answer: $2$

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

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

24

Problem 24: Evaluate $\lim_{x\to2^+}\frac{1}{x-2}$ .

Answer: $+\infty$

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

Identify the dominant behavior: positive denominator approaching zero.
Divide by the highest relevant power or use the sign of the vanishing factor.
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.